Q113 / 160 · AP EAMCET 2022 (08 Jul Shift 1) · Mathematics

$\begin{aligned} & \int(1+x) \log \left(1+x^2\right) d x=\left(x+\frac{x^2}{2}+\frac{1}{2}\right) \\ & \log \left(1+x^2\right)+g(x)+C, \text { then } g(x)=\end{aligned}$

  • Correct answer
    A
    $-2 x-\frac{x^2}{2}+2 \tan ^{-1} x$
  • B
    $2 \tan ^{-1} x+\frac{x^2}{2}+\frac{x^3}{3}$
  • C
    $2 \tan ^{-1} x-\frac{x^2}{2}+3 x$
  • D
    $2 \tan ^{-1} x+3 x+\frac{x^3}{2}$

Solution

Here $\int(1+x) \log \left(1+x^2\right) d x$ $=\int \log \left(1+x^2\right) \cdot 1 d x+\int x \cdot \log \left(1+x^2\right) d x$ $\begin{aligned}=\log \left(1+x^2\right) \cdot x-\int\left(\frac{1}{1+x^2}\right) \cdot( & (2 x) \cdot x d x \\ & +\int x \cdot \log \left(1+x^2\right) d x\end{aligned}$ $=x \log \left(1+x^2\right)-2 \int\left(\frac{x^2+1-1}{1+x^2}\right) d x+\int x \cdot \log \left(1+x^2\right) d x$ $\begin{aligned}=x \log \left(1+x^2\right)-2\left[\int\left(1-\frac{1}{1+x^2}\right) d x\right] & \\ & +\int x \cdot \log \left(1+x^2\right) d x\end{aligned}$ $\begin{aligned} & =x \cdot \log \left(1+x^2\right)-2\left[\int 1 d x-\int\left(\frac{1}{1+x^2}\right) d x\right]+\int x \cdot \log \\ & =x \cdot \log \left(1+x^2\right)-2\left[x-\tan ^{-1} x\right]+\int x \cdot \log \left(1+x^2\right) d x \\ & =x \cdot \log \left(1+x^2\right)-2 x+2 \tan ^{-1} x \int x \cdot \log \left(1+x^2\right) d x\end{aligned}$ Now from $\int \mathrm{x} \cdot \log \left(1+\mathrm{x}^2\right) \mathrm{dx}$ let $1+\mathrm{x}^2=\mathrm{t}$ $2 x d x=d t$ $\begin{aligned} & =\frac{1}{2} \int \log \mathrm{t} d \mathrm{t}=\frac{1}{2}\left[\log \mathrm{t} \cdot \mathrm{t}-\int\left(\frac{\mathrm{t}}{\mathrm{t}} \cdot \mathrm{t}\right) \mathrm{dt}\right] \\ & =-[\mathrm{t} \operatorname{lot} \mathrm{t} \quad \mathrm{t}] \\ & =\frac{1}{2}\left[\left(1+\mathrm{x}^2\right) \log \left(1+\mathrm{x}^2\right)-\left(1+\mathrm{x}^2\right)\right]\end{aligned}$ Now from given integration $\begin{gathered}\int(H x) \log \left(1+x^2\right) d x=x \cdot \log \left(1+x^2\right)-2 x+2 \tan ^{-1} x \\ +\frac{1}{2}\left[\left(1+x^2\right) \log \left(1+x^2\right)-\left(1+x^2\right)\right]+c_1\end{gathered}$ $\begin{array}{r}=\left(\mathrm{x}+\frac{\mathrm{x}^2}{2}+\frac{1}{2}\right) \log \left(1+\mathrm{x}^2\right)-2 \mathrm{x}+2 \tan ^{-1} \mathrm{x}+\frac{1}{2} \\ \left(1+\mathrm{x}^2\right)+\mathrm{c}_1\end{array}$ $=\left(x+\frac{x^2}{2}+\frac{1}{2}\right) \log \left(1+x^2\right)-2 x+2 \tan ^{-1} x-\frac{x^2}{2}$ $+\mathrm{C}, \mathrm{C}=-\frac{1}{2}+\mathrm{C}_1$ $\Rightarrow g(x)=-2 x+2 \tan ^{-1} x-\frac{x^2}{2}$

Hence, the correct answer is option: 'A'.

Relevant Questions