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

Explore the complete step-by-step solutions of the Integration chapter from the SN Dey Mathematics Book for West Bengal Class 12 students. Each problem is solved like a real classroom demonstration — showing every step, applied formula, and concept in a clean, animated layout.

The solutions include exercises 5- Indefinite Integral (Short Answer Type Question) with proper formulas used and clear mathematical formatting. Each answer also carries a small watermark — “Solved by TheMathFellow” — to maintain originality and authenticity.

Topics Covered: Integration of standard functions, substitution method, integration by parts, trigonometric transformations, and definite integrals.

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Integrals — Book Style | myqpaper.in

1(i) \(\displaystyle \int \dfrac{3x^{2}-4x+5}{\sqrt{x}}\,dx\)

\(\dfrac{3x^{2}-4x+5}{\sqrt{x}} = 3x^{3/2}-4x^{1/2}+5x^{-1/2}\)

\(\Rightarrow \int 3x^{3/2}-4x^{1/2}+5x^{-1/2}\,dx = \frac{6}{5}x^{5/2}-\frac{8}{3}x^{3/2}+10x^{1/2}+C\)

\(\boxed{\frac{6}{5}x^{5/2}-\frac{8}{3}x^{3/2}+10x^{1/2}+C}\)

1(ii) \(\displaystyle \int \dfrac{(x^{2}+1)^{2}}{x^{3}}\,dx\)

\((x^{2}+1)^{2}=x^{4}+2x^{2}+1\Rightarrow\dfrac{x^{4}+2x^{2}+1}{x^{3}}=x+\dfrac{2}{x}+\dfrac{1}{x^{3}}\)

\(\Rightarrow \int (x+\dfrac{2}{x}+\dfrac{1}{x^{3}})\,dx = \frac{x^{2}}{2}+2\ln|x|-\frac{1}{2x^{2}}+C\)

\(\boxed{\frac{x^{2}}{2}+2\ln|x|-\frac{1}{2x^{2}}+C}\)

1(iii) \(\displaystyle \int (a^{2/3}+x^{2/3})^{3}\,dx\)

\((a^{2/3}+x^{2/3})^{3} = a^{2}+3a^{4/3}x^{2/3}+3a^{2/3}x^{4/3}+x^{2}\)
\(\Rightarrow \int(a^{2}+3a^{4/3}x^{2/3}+3a^{2/3}x^{4/3}+x^{2})\,dx\)
\(\ = a^{2}x + 3a^{4/3}\int x^{2/3}\,dx + 3a^{2/3}\int x^{4/3}\,dx + \int x^{2}\,dx\)
\(\Rightarrow a^{2}x + 3a^{4/3}\Big(\frac{3x^{5/3}}{5}\Big) + 3a^{2/3}\Big(\frac{3x^{7/3}}{7}\Big) + \frac{x^{3}}{3} + C\)
\(\Rightarrow \boxed{a^{2}x + \frac{9a^{4/3}x^{5/3}}{5} + \frac{9a^{2/3}x^{7/3}}{7} + \frac{x^{3}}{3} + C}\)
\(\boxed{a^{2}x + \frac{9a^{4/3}x^{5/3}}{5} + \frac{9a^{2/3}x^{7/3}}{7} + \frac{x^{3}}{3} + C}\)

1(iv) \(\displaystyle \int \dfrac{12x^{2}-7x-10}{3x+2}\,dx\)

\((3x+2)(4x-5)=12x^{2}-7x-10\Rightarrow\dfrac{12x^{2}-7x-10}{3x+2}=4x-5\)

\(\Rightarrow \int(4x-5)\,dx=2x^{2}-5x+C\)

\(\boxed{2x^{2}-5x+C}\)

1(v) \(\displaystyle \int \dfrac{x^{3}-6x-9}{x-3}\,dx\)

Divide: \(x^{3}-6x-9=(x-3)(x^{2}+3x+3)\)

\(\Rightarrow \dfrac{x^{3}-6x-9}{x-3}=x^{2}+3x+3\)

\(\Rightarrow \int(x^{2}+3x+3)\,dx=\frac{x^{3}}{3}+\frac{3x^{2}}{2}+3x+C\)

\(\boxed{\frac{x^{3}}{3}+\frac{3x^{2}}{2}+3x+C}\)

1(vi) \(\displaystyle \int \dfrac{x^{6}-1}{x-1}\,dx\)

\(\dfrac{x^{6}-1}{x-1}=x^{5}+x^{4}+x^{3}+x^{2}+x+1\)

\(\Rightarrow \int(x^{5}+x^{4}+x^{3}+x^{2}+x+1)\,dx=\frac{x^{6}}{6}+\frac{x^{5}}{5}+\frac{x^{4}}{4}+\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C\)

\(\boxed{\frac{x^{6}}{6}+\frac{x^{5}}{5}+\frac{x^{4}}{4}+\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C}\)

1(vii) \(\displaystyle \int \dfrac{x^{3}-4x^{2}+5x-1}{(x-1)^{2}}\,dx\)

Divide: \(x^{3}-4x^{2}+5x-1=(x-1)^{2}(x-2)+1\)

\(\Rightarrow \dfrac{x^{3}-4x^{2}+5x-1}{(x-1)^{2}}=x-2+\dfrac{1}{(x-1)^{2}}\)

\(\Rightarrow \int(x-2+\dfrac{1}{(x-1)^{2}})\,dx=\frac{x^{2}}{2}-2x-\frac{1}{x-1}+C\)

\(\boxed{\frac{x^{2}}{2}-2x-\frac{1}{x-1}+C}\)

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

\(x^{4}+x^{2}+1=(x^{2}-x+1)(x^{2}+x+1)\)

\(\Rightarrow \dfrac{x^{4}+x^{2}+1}{x^{2}-x+1}=x^{2}+x+1\)

\(\Rightarrow \int(x^{2}+x+1)\,dx=\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C\)

\(\boxed{\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+C}\)

1(ix) \(\displaystyle \int \dfrac{2x^{3}-7x+2}{x^{3}+2x^{2}}\,dx\)

Denominator \(=x^{2}(x+2)\)
\(x=-2\Rightarrow 0\Rightarrow (x+2)\) factor of numerator
Divide \(2x^{3}-7x+2=(x+2)(2x^{2}-4x+1)\)
\(\Rightarrow \dfrac{2x^{3}-7x+2}{x^{2}(x+2)}=\dfrac{2x^{2}-4x+1}{x^{2}}=2-\dfrac{4}{x}+\dfrac{1}{x^{2}}\)
\(\Rightarrow \int(2-\dfrac{4}{x}+\dfrac{1}{x^{2}})\,dx=2x-4\ln|x|-\dfrac{1}{x}+C\)
\(\boxed{2x-4\ln|x|-\dfrac{1}{x}+C}\)

2(i) \(\displaystyle \int \dfrac{e^{6\log x}-e^{4\log x}}{e^{3\log x}-e^{\log x}}\,dx\)

\(e^{k\log x}=x^{k}\Rightarrow\dfrac{x^{6}-x^{4}}{x^{3}-x}=\dfrac{x^{4}(x^{2}-1)}{x(x^{2}-1)}=x^{3}\)

\(\Rightarrow \int x^{3}\,dx=\frac{x^{4}}{4}+C\)

\(\boxed{\frac{x^{4}}{4}+C}\)

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

      \(\dfrac{e^{2x-1}-e^{1-2x}}{e^{x+2}}
      = e^{(2x-1)-(x+2)} - e^{(1-2x)-(x+2)}\)
    
      \(\ = e^{x-3} - e^{-3x-1}\)
    
      \(\displaystyle \int\big(e^{x-3}-e^{-3x-1}\big)\,dx
      = \int e^{x-3}\,dx - \int e^{-3x-1}\,dx\)
    
      \(\displaystyle \int e^{x-3}\,dx = e^{-3}\int e^{x}\,dx = e^{x-3}\)
    
      \(\displaystyle \int e^{-3x-1}\,dx = e^{-1}\int e^{-3x}\,dx
      = e^{-1}\cdot\frac{e^{-3x}}{-3} = -\tfrac{1}{3}e^{-3x-1}\)
    
      \(\Rightarrow \displaystyle \int \frac{e^{2x-1}-e^{1-2x}}{e^{x+2}}\,dx
      = e^{x-3} + \tfrac{1}{3}e^{-3x-1} + C\)
    
      \(\boxed{\,e^{x-3} + \tfrac{1}{3}e^{-3x-1} + C\,}\)

3(ii) \(\displaystyle \int \dfrac{27^{1+x}+9^{1-x}}{3^{x}}\,dx\)

\(27^{1+x}=27\cdot3^{3x},\ 9^{1-x}=9\cdot3^{-2x}\)

\(\Rightarrow \dfrac{27\cdot3^{3x}+9\cdot3^{-2x}}{3^{x}}=27\cdot3^{2x}+9\cdot3^{-3x}\)

\(\Rightarrow \int(27\cdot3^{2x}+9\cdot3^{-3x})\,dx =\frac{27\cdot3^{2x}}{2\ln3}-\frac{3\cdot3^{-3x}}{\ln3}+C\)

\(\boxed{\frac{27\cdot3^{2x}}{2\ln3}-\frac{3\cdot3^{-3x}}{\ln3}+C}\)

3(iii) \(\displaystyle \int \dfrac{2^{6x}-1}{2^{2x}-1}\,dx\)

Let \(t=2^{2x}\Rightarrow2^{6x}=t^{3}\)
\(\Rightarrow\dfrac{t^{3}-1}{t-1}=t^{2}+t+1\)
\(dt=2\ln2\,t\,dx\Rightarrow dx=\dfrac{dt}{2\ln2\,t}\)
\(\Rightarrow \int(t^{2}+t+1)\,dx=\frac{1}{2\ln2}\int(t+1+\tfrac{1}{t})\,dt\)
\(\Rightarrow \frac{1}{2\ln2}(\frac{t^{2}}{2}+t+\ln|t|)+C\)
Substitute \(t=2^{2x}\Rightarrow\boxed{\frac{2^{4x}}{4\ln2}+\frac{2^{2x}}{2\ln2}+x+C}\)
\(\boxed{\frac{2^{4x}}{4\ln2}+\frac{2^{2x}}{2\ln2}+x+C}\)