WB Board Class 12 Mathematics – Integration (SN Dey) Solutions

All 2-marks substitution problems solved step-by-step by The Math Fellow.

Here you get exam-type standard integrals from the Integration chapter, solved in short 2-mark format. Each solution shows the correct substitution and the standard result, just like you would write in the answer script.

This page is currently focused on the 12 important integrals from the image. More 4-marks and 5-marks problems will be added soon.

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Integration — Method of Substitution (2-Marks Set)

1. \( \displaystyle \int \frac{dx}{4x^{2}+9} \)

\(4x^{2}+9 = 4\!\left(x^{2}+\left(\tfrac{3}{2}\right)^{2}\right)\)

\( I = \displaystyle \int \frac{dx}{4x^{2}+9} = \frac{1}{4}\int \frac{dx}{x^{2}+\left(\tfrac{3}{2}\right)^{2}} \)

\( I = \frac{1}{4}\cdot\frac{2}{3}\tan^{-1}\frac{2x}{3} + C \)

\( \displaystyle \frac{1}{6}\tan^{-1}\frac{2x}{3} + C \)

2. \( \displaystyle \int \frac{dx}{2x^{2}-5} \)

\( I = \displaystyle \int \frac{dx}{2x^{2}-5} = \frac{1}{2}\int \frac{dx}{x^{2}-\left(\sqrt{\tfrac{5}{2}}\right)^{2}} \)

\( I = \frac{1}{4\sqrt{\tfrac{5}{2}}} \log\Big|\frac{x-\sqrt{\tfrac{5}{2}}}{x+\sqrt{\tfrac{5}{2}}}\Big| + C \)

\( = \frac{1}{2\sqrt{10}} \log\Big|\frac{\sqrt{2}\,x-\sqrt{5}}{\sqrt{2}\,x+\sqrt{5}}\Big| + C \)

\( \displaystyle \frac{1}{2\sqrt{10}} \log\Big|\frac{\sqrt{2}\,x-\sqrt{5}}{\sqrt{2}\,x+\sqrt{5}}\Big| + C \)

3. \( \displaystyle \int \frac{dx}{16-25x^{2}} \)

\( 16-25x^{2}=4^{2}-(5x)^{2}. \)

\( I = \frac{1}{2\cdot4\cdot5} \log\Big|\frac{4+5x}{4-5x}\Big| + C \)

\( \displaystyle \frac{1}{40} \log\Big|\frac{4+5x}{4-5x}\Big| + C \)

6. \( \displaystyle \int \frac{dx}{\sqrt{x}\,(4x-49)} \)

Put \( t=\sqrt{x}\Rightarrow x=t^{2},\;dx=2t\,dt. \)

\( I = \displaystyle \int \frac{2t\,dt}{t(4t^{2}-49)} = 2\int\frac{dt}{4t^{2}-49} = \frac{1}{2}\int\frac{dt}{t^{2}-\left(\tfrac{7}{2}\right)^{2}} \)

\( I = \frac{1}{4\cdot \tfrac{7}{2}} \log\Big|\frac{t-\tfrac{7}{2}}{t+\tfrac{7}{2}}\Big|+C = \frac{1}{14}\log\Big|\frac{2t-7}{2t+7}\Big|+C \)

\( \displaystyle \frac{1}{14} \log\Big|\frac{2\sqrt{x}-7}{2\sqrt{x}+7}\Big| + C \)

7. \( \displaystyle \int \frac{dx}{x\sqrt{x+1}} \)

Put \( t=\sqrt{x+1}\Rightarrow x=t^{2}-1,\;dx=2t\,dt. \)

\( I = \displaystyle \int \frac{2t\,dt}{(t^{2}-1)t} = 2\int\frac{dt}{t^{2}-1} \)

\( \displaystyle \frac{2}{t^{2}-1} = \frac{1}{t-1}-\frac{1}{t+1}. \)

\( I = \int\Big(\frac{1}{t-1}-\frac{1}{t+1}\Big)\,dt = \log|t-1|-\log|t+1|+C \)

\( \displaystyle \log\Big|\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\Big| + C \)

8. \( \displaystyle \int \frac{dx}{e^{x}+e^{-x}} \) [NCERT]

\( I = \displaystyle \int \frac{dx}{e^{x}+e^{-x}} = \int \frac{e^{x}\,dx}{e^{2x}+1}. \)

Put \( t=e^{x}\Rightarrow dt=e^{x}\,dx. \)

\( I = \displaystyle \int \frac{dt}{t^{2}+1} = \tan^{-1}t + C \)

\( \displaystyle \tan^{-1}(e^{x}) + C \)

9. \( \displaystyle \int \frac{dx}{\sqrt{x}+x\sqrt{x}} \)

Denominator \(=\sqrt{x}(1+x).\) Put \( t=\sqrt{x}\Rightarrow x=t^{2},\;dx=2t\,dt. \)

\( I = \displaystyle \int \frac{2t\,dt}{t(1+t^{2})} = 2\int\frac{dt}{1+t^{2}} \)

\( I = 2\tan^{-1}t + C \)

\( \displaystyle 2\tan^{-1}(\sqrt{x}) + C \)

10. \( \displaystyle \int \frac{dx}{\sqrt{16x^{2}+25}} \)

Put \( u = 4x \Rightarrow du = 4\,dx,\; dx = \dfrac{du}{4}. \)

\( I = \displaystyle \frac{1}{4}\int \frac{du}{\sqrt{u^{2}+25}} \)

\( I = \frac{1}{4} \log\big|u+\sqrt{u^{2}+25}\big|+C \)

\( \displaystyle \frac{1}{4} \log\big|4x+\sqrt{16x^{2}+25}\big|+C \)

11. \( \displaystyle \int \frac{dx}{\sqrt{3x^{2}-5}} \)

Put \( u=\sqrt{3}\,x \Rightarrow du=\sqrt{3}\,dx,\;dx=\dfrac{du}{\sqrt{3}}. \)

\( I = \displaystyle \frac{1}{\sqrt{3}} \int \frac{du}{\sqrt{u^{2}-5}} \)

\( I = \frac{1}{\sqrt{3}} \log\big|u+\sqrt{u^{2}-5}\big|+C \)

\( \displaystyle \frac{1}{\sqrt{3}} \log\big(\sqrt{3}\,x+\sqrt{3x^{2}-5}\big)+C \)

12. \( \displaystyle \int \frac{dx}{\sqrt{49-9x^{2}}} \)

Put \( u=3x \Rightarrow du=3\,dx,\;dx=\dfrac{du}{3}. \)

\( I = \displaystyle \frac{1}{3} \int \frac{du}{\sqrt{49-u^{2}}} \)

\( I = \frac{1}{3}\sin^{-1}\frac{u}{7}+C \)

\( \displaystyle \frac{1}{3}\sin^{-1}\frac{3x}{7}+C \)

13. \( \displaystyle \int \dfrac{dx}{x + x(\log x)^{2}} \)

\( \displaystyle I = \int \dfrac{dx}{x\big(1+(\log x)^{2}\big)} \)

Put \(t=\log x,\; dt=\dfrac{dx}{x}.\)

\( I = \displaystyle \int \dfrac{dt}{1+t^{2}} \)

\( I = \tan^{-1} t + C \)

\( \tan^{-1}(\log x) + C \)

14. \( \displaystyle \int \dfrac{e^{x}\,dx}{\sqrt{e^{2x}-a^{2}}} \)

Put \(u=e^{x},\; du=e^{x}dx.\)

\( I = \displaystyle \int \dfrac{du}{\sqrt{u^{2}-a^{2}}} \)

\( I = \log\big|u+\sqrt{u^{2}-a^{2}}\big| + C \)

\( \log\big|e^{x}+\sqrt{e^{2x}-a^{2}}\big| + C \)

15. \( \displaystyle \int \dfrac{\cos x\,dx}{8+\cos^{2}x} \)

Put \(t=\sin x,\; dt=\cos x\,dx.\)

\( 8+\cos^{2}x = 8 + (1-\sin^{2}x) = 9-t^{2}. \)

\( I = \displaystyle \int \dfrac{dt}{9-t^{2}} \)

\( I = \dfrac{1}{6}\log\left|\dfrac{3+t}{3-t}\right|+C \)

\( \dfrac{1}{6}\log\left|\dfrac{3+\sin x}{3-\sin x}\right|+C \)

16. \( \displaystyle \int \dfrac{d(x^{2}-1)}{x\sqrt{2-x^{2}}} \)

\( d(x^{2}-1)=2x\,dx \Rightarrow I = \displaystyle \int \dfrac{2x\,dx}{x\sqrt{2-x^{2}}} = 2\int \dfrac{dx}{\sqrt{2-x^{2}}}. \)

Put \(u=\dfrac{x}{\sqrt{2}},\; dx=\sqrt{2}\,du.\)

\( I = 2\sqrt{2}\int \dfrac{du}{\sqrt{2(1-u^{2})}} = 2\int \dfrac{du}{\sqrt{1-u^{2}}} = 2\sin^{-1}u + C. \)

\( 2\sin^{-1}\!\left(\dfrac{x}{\sqrt{2}}\right) + C \)

17. \( \displaystyle \int \dfrac{x^{2}+\sin^{2}x}{1+x^{2}}\sec^{2}x\,dx \)

\( \dfrac{x^{2}+\sin^{2}x}{1+x^{2}} = 1-\dfrac{\cos^{2}x}{1+x^{2}}. \)

\( I = \displaystyle \int \sec^{2}x\,dx - \int \dfrac{\sec^{2}x\cos^{2}x}{1+x^{2}}\,dx \)

\( I = \int \sec^{2}x\,dx - \int \dfrac{dx}{1+x^{2}} \)

\( I = \tan x - \tan^{-1}x + C \)

\( \tan x - \tan^{-1}x + C \)

18. \( \displaystyle \int \dfrac{d(\log x)}{x+1} \)

\( d(\log x)=\dfrac{dx}{x} \Rightarrow I = \displaystyle \int \dfrac{dx}{x(x+1)}. \)

\( \dfrac{1}{x(x+1)}=\dfrac{1}{x}-\dfrac{1}{x+1}. \)

\( I = \log|x|-\log|x+1|+C. \)

\( \log\left|\dfrac{x}{x+1}\right|+C \)

19. \( \displaystyle \int \dfrac{1+2x^{2}}{x^{2}(1+x^{2})}\,dx \)

\( \displaystyle \dfrac{1+2x^{2}}{x^{2}(1+x^{2})} = \dfrac{1}{x^{2}} + \dfrac{1}{1+x^{2}}. \)

\( I = \displaystyle \int \dfrac{dx}{x^{2}} + \int \dfrac{dx}{1+x^{2}} \)

\( I = -\dfrac{1}{x} + \tan^{-1}x + C \)

\( -\dfrac{1}{x} + \tan^{-1}x + C \)

20. \( \displaystyle \int \dfrac{\cos x\,dx}{\sqrt{\cos2x}} \)

Put \(t=\sin x,\; dt=\cos x\,dx.\)

            \( \cos2x = 1-2\sin^{2}x = 1-2t^{2}. \)
          
            \( I = \displaystyle \int \dfrac{dt}{\sqrt{1-2t^{2}}}. \)
          
            Put \(u=\sqrt{2}\,t,\; du=\sqrt{2}\,dt.\)
          
            \( I = \dfrac{1}{\sqrt{2}}\int \dfrac{du}{\sqrt{1-u^{2}}}
               = \dfrac{1}{\sqrt{2}}\sin^{-1}u + C. \)
          
              \( \dfrac{1}{\sqrt{2}}\sin^{-1}\!\big(\sqrt{2}\sin x\big) + C \)

21. \( \displaystyle \int \dfrac{\sin x\,dx}{\cos2x} \)

Put \(t=\cos x,\; dt=-\sin x\,dx.\)

            \( \cos2x = 2\cos^{2}x-1 = 2t^{2}-1. \)
          
            \( I = -\displaystyle \int \dfrac{dt}{2t^{2}-1}
               = \int \dfrac{dt}{1-2t^{2}}. \)
          
            Put \(u=\sqrt{2}\,t,\; du=\sqrt{2}\,dt.\)
          
            \( I = \dfrac{1}{\sqrt{2}}\int \dfrac{du}{1-u^{2}}
               = \dfrac{1}{2\sqrt{2}}
                 \log\left|\dfrac{1+u}{1-u}\right| + C. \)
          
              \( \dfrac{1}{2\sqrt{2}}
                 \log\left|\dfrac{1+\sqrt{2}\cos x}{1-\sqrt{2}\cos x}\right|
                 + C \)

22. \( \displaystyle \int \dfrac{dx}{\sqrt{1-x^{2}}\{1+(\sin^{-1}x)^{2}\}} \)

Put \(t=\sin^{-1}x,\; dt=\dfrac{dx}{\sqrt{1-x^{2}}}.\)

\( I = \displaystyle \int \dfrac{dt}{1+t^{2}} = \tan^{-1}t + C. \)

\( \tan^{-1}\!\big(\sin^{-1}x\big) + C \)

23. \( \displaystyle \int \dfrac{dx}{x\sqrt{4-(\log x)^{2}}} \)

Put \(t=\log x,\; dt=\dfrac{dx}{x}.\)

\( I = \displaystyle \int \dfrac{dt}{\sqrt{4-t^{2}}} = \sin^{-1}\!\left(\dfrac{t}{2}\right)+C. \)

\( \sin^{-1}\!\left(\dfrac{\log x}{2}\right)+C \)

24. \( \displaystyle \int \dfrac{dx}{x^{2}-2ax} \)

            \( x^{2}-2ax = (x-a)^{2}-a^{2}. \)
          
            \( \displaystyle
              I = \int \dfrac{dx}{(x-a)^{2}-a^{2}}
              = \int \dfrac{dx}{a^{2}\left[\left(\dfrac{x-a}{a}\right)^{2}-1\right]}.
            \)
          
            Put \( t = \dfrac{x-a}{a} \Rightarrow dx = a\,dt. \)
          
            \( \displaystyle
              I = \dfrac{1}{a}\int \dfrac{dt}{t^{2}-1}
              = \dfrac{1}{a}\cdot\dfrac{1}{2}\ln\left|\dfrac{t-1}{t+1}\right|+C.
            \)
          
            \( t-1 = \dfrac{x-a}{a}-1=\dfrac{x-2a}{a},\quad
               t+1 = \dfrac{x-a}{a}+1=\dfrac{x}{a}. \)
          
              \( \displaystyle I
              = \dfrac{1}{2a}\ln\left|\dfrac{x-2a}{x}\right|+C \)

25. \( \displaystyle \int \dfrac{dx}{2ax-x^{2}} \)

            \( 2ax-x^{2} = a^{2}-(x-a)^{2}. \)
          
            \( \displaystyle
              I = \int \dfrac{dx}{a^{2}-(x-a)^{2}}
              = \int \dfrac{dx}{a^{2}\left[1-\left(\dfrac{x-a}{a}\right)^{2}\right]}.
            \)
          
            Put \( t = \dfrac{x-a}{a} \Rightarrow dx = a\,dt. \)
          
            \( \displaystyle
              I = \dfrac{1}{a}\int \dfrac{dt}{1-t^{2}}
              = \dfrac{1}{a}\cdot\dfrac{1}{2}
                \ln\left|\dfrac{1+t}{1-t}\right|+C.
            \)
          
            \( 1+t = 1+\dfrac{x-a}{a}=\dfrac{x}{a},\quad
               1-t = 1-\dfrac{x-a}{a}=\dfrac{2a-x}{a}. \)
          
              \( \displaystyle I
              = \dfrac{1}{2a}\ln\left|\dfrac{x}{2a-x}\right|+C \)

4-marks substitution questions will be added here soon.

1. \( \displaystyle \int \dfrac{dx}{\sqrt{x^{2}-ax}} \)

\( I = \displaystyle \int \dfrac{dx}{\sqrt{x^{2}-ax}} \)

\( x^{2}-ax = x(x-a). \)

\( I = \displaystyle \int \dfrac{dx}{\sqrt{x(x-a)}} \)

Put \( x = t^{2} \Rightarrow dx = 2t\,dt. \)

\( I = 2\displaystyle \int \dfrac{dt}{\sqrt{t^{2}-a}} \)

\( = 2\log\left|t+\sqrt{t^{2}-a}\right|+C \)

\( = 2\log\left|\sqrt{x}+\sqrt{x-a}\right|+C \)

\( \displaystyle I = 2\log\left|\sqrt{x}+\sqrt{x-a}\right|+C \)

1. \( \displaystyle \int \dfrac{dx}{x^{2}-ax} \)

\( x^{2}-ax = x^{2}-2x\frac{a}{2} =\left(x-\frac{a

2. \( \displaystyle \int \dfrac{dx}{\sqrt{2x-x^{2}}} \)

\(2x-x^{2}=-(x^{2}-2x) =-(x^{2}-2x+1-1)=1-(x-1)^{2}.\)

Put \(t=x-1,\;dt=dx.\)

\(I=\displaystyle\int \dfrac{dt}{\sqrt{1-t^{2}}} =\sin^{-1}t+C=\sin^{-1}(x-1)+C.\)

\( \sin^{-1}(x-1)+C \)

3. \( \displaystyle \int \dfrac{\sec^{2}x\,dx}{\sqrt{5-\sec^{2}x}} \)

\(5-\sec^{2}x=5-(1+\tan^{2}x) =4-\tan^{2}x.\)

Put \(t=\tan x,\;dt=\sec^{2}x\,dx.\)

\(I=\displaystyle\int \dfrac{dt}{\sqrt{4-t^{2}}} =\sin^{-1}\!\left(\frac{t}{2}\right)+C =\sin^{-1}\!\left(\frac{\tan x}{2}\right)+C.\)

\( \sin^{-1}\!\left(\dfrac{\tan x}{2}\right)+C \)

4. \( \displaystyle \int \dfrac{dx}{x^{2}+4x+8} \)

\(x^{2}+4x+8=x^{2}+4x+4+4 =(x+2)^{2}+4.\)

Put \(t=x+2,\;dt=dx.\)

\(I=\displaystyle\int \dfrac{dt}{t^{2}+4} =\frac{1}{2}\tan^{-1}\!\left(\frac{t}{2}\right)+C.\)

\( \dfrac{1}{2}\tan^{-1}\!\left(\dfrac{x+2}{2}\right)+C \)

5. \( \displaystyle \int \dfrac{dx}{6x^{2}+7x+2} \)

\(6x^{2}+7x+2=(3x+2)(2x+1).\)

\(\displaystyle \frac{1}{6x^{2}+7x+2} =-\frac{3}{3x+2}+\frac{2}{2x+1}.\)

\(I=-\int\frac{3\,dx}{3x+2}+\int\frac{2\,dx}{2x+1} =-\log|3x+2|+\log|2x+1|+C.\)

\( \log\left|\dfrac{2x+1}{3x+2}\right|+C \)

6. \( \displaystyle \int \dfrac{dx}{1+x-x^{2}} \)

\(1+x-x^{2}=-(x^{2}-x-1).\)

\(x^{2}-x-1=\left(x-\tfrac12\right)^{2} -\left(\tfrac{\sqrt5}{2}\right)^{2}.\)

\(\Rightarrow 1+x-x^{2} =-\Big[(x-\tfrac12)^{2}-\big(\tfrac{\sqrt5}{2}\big)^{2}\Big].\)

\(I=-\displaystyle\int \dfrac{dx}{(x-\tfrac12)^{2}-\big(\tfrac{\sqrt5}{2}\big)^{2}} =\frac{1}{\sqrt5} \log\left|\frac{x-\tfrac12+\tfrac{\sqrt5}{2}} {x-\tfrac12-\tfrac{\sqrt5}{2}}\right|+C.\)

\( \dfrac{1}{\sqrt5} \log\left|\dfrac{x-\tfrac12+\tfrac{\sqrt5}{2}} {x-\tfrac12-\tfrac{\sqrt5}{2}}\right|+C \)

7(i). \( \displaystyle \int \dfrac{x\,dx}{4x^{4}+4x^{2}+5} \)

Put \(t=2x^{2}+1\Rightarrow dt=4x\,dx,\;x\,dx=\dfrac{dt}{4}.\)

\(4x^{4}+4x^{2}+5=(2x^{2}+1)^{2}+4=t^{2}+4.\)

\(I=\dfrac{1}{4}\displaystyle\int \dfrac{dt}{t^{2}+4} =\dfrac{1}{8}\tan^{-1}\!\left(\frac{t}{2}\right)+C.\)

\( \dfrac{1}{8}\tan^{-1}\!\left(\dfrac{2x^{2}+1}{2}\right)+C \)

7(ii). \( \displaystyle \int \dfrac{x\,dx}{x^{4}-x^{2}+1} \)

Put \(t=x^{2}-\tfrac12\Rightarrow dt=2x\,dx,\;x\,dx=\dfrac{dt}{2}.\)

\(x^{4}-x^{2}+1=t^{2}+\dfrac{3}{4} =t^{2}+\left(\tfrac{\sqrt3}{2}\right)^{2}.\)

\(I=\dfrac{1}{2}\int\dfrac{dt}{t^{2}+(\tfrac{\sqrt3}{2})^{2}} =\dfrac{1}{\sqrt3}\tan^{-1}\!\left(\dfrac{2t}{\sqrt3}\right)+C.\)

\( \dfrac{1}{\sqrt3} \tan^{-1}\!\left(\dfrac{2x^{2}-1}{\sqrt3}\right)+C \)

8. \( \displaystyle \int \dfrac{e^{x}\,dx}{5-4e^{x}-e^{2x}} \)

Put \(t=e^{x},\;dt=e^{x}dx.\)

\(5-4t-t^{2}=-(t^{2}+4t-5)=-(t-1)(t+5).\)

\(\displaystyle -\frac{1}{(t-1)(t+5)} =-\frac{1}{6}\cdot\frac{1}{t-1} +\frac{1}{6}\cdot\frac{1}{t+5}.\)

\(I=\dfrac{1}{6}\log\left|\dfrac{t+5}{t-1}\right|+C =\dfrac{1}{6}\log\left|\dfrac{e^{x}+5}{e^{x}-1}\right|+C.\)

\( \dfrac{1}{6}\log\left|\dfrac{e^{x}+5}{e^{x}-1}\right|+C \)

9. \( \displaystyle \int \dfrac{dx}{x\sqrt{1+x^{3}}} \)

Put \(t=\sqrt{1+x^{3}}\Rightarrow t^{2}=1+x^{3}.\)

\(2t\,dt=3x^{2}dx\Rightarrow dx=\dfrac{2t}{3x^{2}}dt.\)

\(I=\displaystyle\int \dfrac{1}{x\,t}\cdot\dfrac{2t}{3x^{2}}dt =\dfrac{2}{3}\int \dfrac{dt}{x^{3}} =\dfrac{2}{3}\int \dfrac{dt}{t^{2}-1}.\)

\(\displaystyle I=\dfrac{1}{3} \log\left|\dfrac{t-1}{t+1}\right|+C =\dfrac{1}{3}\log\left|\dfrac{\sqrt{1+x^{3}}-1} {\sqrt{1+x^{3}}+1}\right|+C.\)

\( \dfrac{1}{3}\log\left|\dfrac{\sqrt{1+x^{3}}-1} {\sqrt{1+x^{3}}+1}\right|+C \)

10. \( \displaystyle \int \dfrac{dx}{x[(\log x)^{2}-6\log x+5]} \)

Put \(t=\log x,\;dt=\dfrac{dx}{x}.\)

\(t^{2}-6t+5=(t-1)(t-5).\)

\(\displaystyle \frac{1}{(t-1)(t-5)} =-\frac{1}{4}\cdot\frac{1}{t-1} +\frac{1}{4}\cdot\frac{1}{t-5}.\)

\(I=\dfrac{1}{4} \log\left|\dfrac{t-5}{t-1}\right|+C =\dfrac{1}{4} \log\left|\dfrac{\log x-5}{\log x-1}\right|+C.\)

\( \dfrac{1}{4} \log\left|\dfrac{\log x-5}{\log x-1}\right|+C \)

11. \( \displaystyle \int \dfrac{(2x+2)\,dx}{x^{2}-3x+2} \) or \( \displaystyle \int \dfrac{(x+2)\,dx}{x^{2}-3x+2} \)

\(x^{2}-3x+2=(x-1)(x-2).\)

\(\displaystyle \frac{x+2}{x^{2}-3x+2} =\frac{-3}{x-1}+\frac{4}{x-2}.\)

\(I=-3\log|x-1|+4\log|x-2|+C.\)

\( -3\log|x-1|+4\log|x-2|+C \)

12. \( \displaystyle \int \dfrac{(x+2)\,dx}{2x^{2}+6x+5} \)

Write \(x+2=\tfrac12(2x+3)+\tfrac12.\)

\(I=\dfrac12\int\dfrac{2x+3}{2x^{2}+6x+5}dx +\dfrac12\int\dfrac{dx}{2x^{2}+6x+5}.\)

First part \(=\dfrac14\log(2x^{2}+6x+5).\)

\(2x^{2}+6x+5 =2\left[x^{2}+3x+\tfrac{9}{4}\right]-\tfrac{9}{2}+5 =2\Big(x+\tfrac32\Big)^{2}+\tfrac12.\)

\(\displaystyle \int\dfrac{dx}{2x^{2}+6x+5} =\tan^{-1}(2x+3).\)

\(I=\dfrac14\log(2x^{2}+6x+5)+\dfrac12\tan^{-1}(2x+3)+C.\)

\( \dfrac14\log(2x^{2}+6x+5) +\dfrac12\tan^{-1}(2x+3)+C \ )

13. \( \displaystyle \int \dfrac{(2x+1)\,dx}{x(x+3)} \)

\(\displaystyle \frac{2x+1}{x(x+3)} =\frac{1}{3}\cdot\frac1x+\frac{5}{3}\cdot\frac1{x+3}.\)

\(I=\frac{1}{3}\log|x|+\frac{5}{3}\log|x+3|+C.\)

\( \dfrac{1}{3}\log|x|+\dfrac{5}{3}\log|x+3|+C \)

14. \( \displaystyle \int \dfrac{x^{2}\,dx}{x^{2}-a^{2}} \)

\(\displaystyle \frac{x^{2}}{x^{2}-a^{2}} =1+\frac{a^{2}}{x^{2}-a^{2}}.\)

\(\displaystyle \frac{1}{x^{2}-a^{2}} =\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right).\)

\(I=\int dx+\frac{a^{2}}{2a} \int\left(\frac{1}{x-a}-\frac{1}{x+a}\right)dx\) \(=x+\frac{a}{2}\log\left|\frac{x-a}{x+a}\right|+C.\)

\( x+\dfrac{a}{2}\log\left|\dfrac{x-a}{x+a}\right|+C \)

15. \( \displaystyle \int \dfrac{x^{2}\,dx}{x^{2}+6x+12} \)

\(\displaystyle \frac{x^{2}}{x^{2}+6x+12} =1-\frac{6x+12}{x^{2}+6x+12}.\)

\(I=\int dx-\int\dfrac{6x+12}{x^{2}+6x+12}dx.\)

Put \(t=x^{2}+6x+12,\;dt=(2x+6)dx.\)

\(\displaystyle \int\dfrac{6x+12}{t}dx =3\int\dfrac{2x+6}{t}dx=3\log t.\)

\(x^{2}+6x+12 =(x^{2}+6x+9)+3=(x+3)^{2}+3.\)

\(I=x-3\log(x^{2}+6x+12) +2\sqrt3\,\tan^{-1}\!\left(\dfrac{x+3}{\sqrt3}\right)+C.\)

\( x-3\log(x^{2}+6x+12) +2\sqrt3\,\tan^{-1}\!\left(\dfrac{x+3}{\sqrt3}\right)+C \)

16. \( \displaystyle \int \dfrac{16x^{4}+2x+2}{4x^{2}+1}\,dx \)

Divide: \(\displaystyle \frac{16x^{4}+2x+2}{4x^{2}+1} =4x^{2}-1+\frac{2x+3}{4x^{2}+1}.\)

\(I=\int(4x^{2}-1)dx+\int\dfrac{2x+3}{4x^{2}+1}dx.\)

\(2x+3 = \tfrac14\cdot 8x +3.\)

\(\displaystyle I=\frac{4x^{3}}{3}-x+\frac14\int\dfrac{8x}{4x^{2}+1}dx +3\int\dfrac{dx}{4x^{2}+1}.\)

\(=\frac{4x^{3}}{3}-x+\frac14\log(4x^{2}+1) +\frac{3}{2}\tan^{-1}(2x)+C.\)

\( \dfrac{4x^{3}}{3}-x+\dfrac14\log(4x^{2}+1) +\dfrac{3}{2}\tan^{-1}(2x)+C \)

17. \( \displaystyle \int \dfrac{dx}{\sqrt{x^{2}-x-6}} \)

\(x^{2}-x-6 =x^{2}-x+\tfrac14-\tfrac14-6 =\left(x-\tfrac12\right)^{2}-\left(\tfrac52\right)^{2}.\)

Put \(u=x-\tfrac12,\;du=dx.\)

\(I=\displaystyle\int\dfrac{du}{\sqrt{u^{2}-\big(\tfrac52\big)^{2}}} =\log\Big|u+\sqrt{u^{2}-\big(\tfrac52\big)^{2}}\Big|+C.\)

\( \log\Big|x-\tfrac12+\sqrt{x^{2}-x-6}\Big|+C \)

18. \( \displaystyle \int \dfrac{dx}{\sqrt{1-x-x^{2}}} \)

\(1-x-x^{2}=-(x^{2}+x-1).\)

\(x^{2}+x-1 =\left(x+\tfrac12\right)^{2}-\left(\tfrac{\sqrt5}{2}\right)^{2}.\)

\(\Rightarrow 1-x-x^{2} =\dfrac54-\left(x+\tfrac12\right)^{2}.\)

Put \(u=x+\tfrac12,\;du=dx.\)

\(I=\displaystyle\int \dfrac{du}{\sqrt{\big(\tfrac{\sqrt5}{2}\big)^{2}-u^{2}}} =\sin^{-1}\!\left(\frac{2u}{\sqrt5}\right)+C.\)

\( \sin^{-1}\!\left(\dfrac{2x+1}{\sqrt5}\right)+C \)

19. \( \displaystyle \int \dfrac{dx}{\sqrt{5-4x-2x^{2}}} \)

\(5-4x-2x^{2} =-2\Big(x^{2}+2x-\tfrac52\Big).\)

\(x^{2}+2x-\tfrac52 =(x+1)^{2}-\tfrac{7}{2}.\)

\(\Rightarrow 5-4x-2x^{2} =7-2(x+1)^{2}.\)

Put \(u=(x+1)\sqrt{\dfrac{2}{7}},\;dx=\sqrt{\dfrac{7}{2}}\,du.\)

\(I=\dfrac{1}{\sqrt2}\int\dfrac{du}{\sqrt{1-u^{2}}} =\dfrac{1}{\sqrt2}\sin^{-1}u+C.\)

\( \dfrac{1}{\sqrt2}\sin^{-1}\!\left( \dfrac{\sqrt2(x+1)}{\sqrt7}\right)+C \)

20. \( \displaystyle \int \dfrac{dx}{\sqrt{x^{2}-8x+15}} \)

\(x^{2}-8x+15=(x^{2}-8x+16)-1 =(x-4)^{2}-1.\)

Put \(u=x-4,\;du=dx.\)

\(I=\displaystyle\int\dfrac{du}{\sqrt{u^{2}-1}} =\log|u+\sqrt{u^{2}-1}|+C.\)

\( \log\big|x-4+\sqrt{x^{2}-8x+15}\big|+C \)

21. \( \displaystyle \int \dfrac{dx}{\sqrt{(x-\alpha)(x-\beta)}} \)

\((x-\alpha)(x-\beta) =x^{2}-(\alpha+\beta)x+\alpha\beta.\)

Complete square in \(x\) and then use standard form.

\(I=\log\Big|\sqrt{x-\alpha}+\sqrt{x-\beta}\Big|+C.\)

\( \log\Big|\sqrt{x-\alpha}+\sqrt{x-\beta}\Big|+C \)

22. \( \displaystyle \int \dfrac{dx}{\sqrt{5x^{2}+6x+4}} \)

Put \(u=x+\dfrac{3}{5},\;du=dx.\)

\(5x^{2}+6x+4 =5\left(x^{2}+\dfrac{6}{5}x\right)+4 =5\left[(x+\tfrac35)^{2}-\tfrac{9}{25}\right]+4 =5u^{2}+\dfrac{11}{5}.\)

\(I=\sqrt5\int\dfrac{du}{\sqrt{25u^{2}+11}} =\dfrac{\sqrt5}{5}\log\Big|u+\sqrt{u^{2}+\tfrac{11}{25}}\Big|+C.\)

\( \dfrac{\sqrt5}{5}\log\Big|x+\tfrac35 +\sqrt{(x+\tfrac35)^{2}+\tfrac{11}{25}}\Big|+C \)

23. \( \displaystyle \int \dfrac{x\,dx}{\sqrt{(x^{2}-a^{2})(b^{2}-x^{2})}},\quad b^{2}>a^{2} \)

Put \(t=x^{2}.\Rightarrow dt=2x\,dx,\;x\,dx=\dfrac{dt}{2}.\)

\(I=\dfrac{1}{2}\int \dfrac{dt}{\sqrt{(t-a^{2})(b^{2}-t)}}.\)

Put \( t = \dfrac{a^{2}+b^{2}}{2} +\dfrac{b^{2}-a^{2}}{2}\sin\theta,\;dt=\dfrac{b^{2}-a^{2}}{2}\cos\theta\,d\theta.\)

Then \(I=\dfrac{1}{2}\int d\theta =\dfrac{1}{2}\theta +C.\)

\( \sin\theta = \dfrac{2t-a^{2}-b^{2}}{b^{2}-a^{2}} =\dfrac{2x^{2}-a^{2}-b^{2}}{b^{2}-a^{2}}.\)

\( \displaystyle I=\frac{1}{2}\sin^{-1}\!\left( \dfrac{2x^{2}-a^{2}-b^{2}}{b^{2}-a^{2}}\right)+C \)

24. \( \displaystyle \int \dfrac{\sec^{2}x\,dx}{\sqrt{5\sec^{2}x-12\tan x-1}} \)

Put \(t=\tan x,\;dt=\sec^{2}x\,dx.\)

\(5\sec^{2}x-12\tan x-1 =5(1+t^{2})-12t-1 =5t^{2}-12t+4.\)

\(5t^{2}-12t+4 =5\left[t^{2}-\dfrac{12}{5}t\right]+4 =5\Big(t-\tfrac65\Big)^{2}-\dfrac{16}{5}.\)

Put \(u=t-\dfrac65.\)

\(I=\dfrac{1}{\sqrt5}\int \dfrac{du}{\sqrt{u^{2}-\big(\tfrac{4}{\sqrt5}\big)^{2}}} =\dfrac{1}{\sqrt5} \log\Big|u+\sqrt{u^{2}-\big(\tfrac{4}{\sqrt5}\big)^{2}}\Big|+C.\)

\( \dfrac{1}{\sqrt5} \log\Big|\tan x-\tfrac65 +\sqrt{\tan^{2}x-\tfrac{12}{5}\tan x+\tfrac{4}{5}}\Big| +C \)

25. \( \displaystyle \int \dfrac{\cos x\,dx}{\sqrt{6+11\sin x-10\sin^{2}x}} \)

Put \(s=\sin x,\;ds=\cos x\,dx.\)

\(6+11s-10s^{2} =-10\left[s^{2}-\frac{11}{10}s-\frac{3}{5}\right].\)

\(s^{2}-\frac{11}{10}s-\frac{3}{5} =\Big(s-\frac{11}{20}\Big)^{2}-\Big(\frac{19}{20}\Big)^{2}.\)

\(\Rightarrow 6+11s-10s^{2} =10\Big[\Big(\frac{19}{20}\Big)^{2} -\Big(s-\frac{11}{20}\Big)^{2}\Big].\)

Put \(u=s-\dfrac{11}{20}.\)

\(I=\dfrac{1}{\sqrt{10}} \int\dfrac{du}{\sqrt{\big(\tfrac{19}{20}\big)^{2}-u^{2}}} =\dfrac{1}{\sqrt{10}} \sin^{-1}\!\left(\dfrac{20u}{19}\right)+C.\)

\( \dfrac{1}{\sqrt{10}} \sin^{-1}\!\left(\dfrac{20\sin x-11}{19}\right)+C \)

26. \( \displaystyle \int \dfrac{e^{x}\,dx}{\sqrt{e^{2x}-5e^{x}+6}} \)

Put \(t=e^{x},\;dt=e^{x}dx.\)

\(e^{2x}-5e^{x}+6=t^{2}-5t+6.\)

\(t^{2}-5t+6 =(t-\tfrac52)^{2}-\left(\tfrac12\right)^{2}.\)

\(I=\displaystyle\int \dfrac{dt}{\sqrt{(t-\tfrac52)^{2}-\big(\tfrac12\big)^{2}}} =\log\Big|t-\tfrac52+\sqrt{t^{2}-5t+6}\Big|+C.\)

\( \log\Big|e^{x}-\tfrac52+\sqrt{e^{2x}-5e^{x}+6}\Big|+C \)

27. \( \displaystyle \int \dfrac{2^{x}\,dx}{\sqrt{4^{x}-2^{x}+5}} \)

Put \(t=2^{x}\Rightarrow dt=t\ln2\,dx,\;2^{x}dx=\dfrac{dt}{\ln2}.\)

\(4^{x}-2^{x}+5=t^{2}-t+5.\)

\(t^{2}-t+5 =(t-\tfrac12)^{2}+\dfrac{19}{4}.\)

\(I=\dfrac{1}{\ln2} \int\dfrac{dt}{\sqrt{(t-\tfrac12)^{2} +\big(\tfrac{\sqrt{19}}{2}\big)^{2}}} =\dfrac{1}{\ln2} \log\Big|t-\tfrac12+\sqrt{t^{2}-t+5}\Big|+C.\)

\( \dfrac{1}{\ln2} \log\Big|2^{x}-\tfrac12+\sqrt{4^{x}-2^{x}+5}\Big|+C \)

5-marks substitution questions will be added here soon.