Q110 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics
If $\int \frac{\log \left(1+x^4\right)}{x^3} d x=f(x) \log \left(\frac{1}{g(x)}\right)+\tan ^{-1}(h(x))+c$, then $h(x)\left[f(x)+f\left(\frac{1}{x}\right)\right]=$
- A$h(x) g(-x)$
- Correct answerB$\frac{g(x)}{2}$
- C$g(x)+g(-x)$
- D$g(x) h(x)$
Solution
$\begin{aligned}
& \text { Given, } \int \frac{\log \left(1+x^4\right)}{x^3} d x \\
= & f(x) \log \left(\frac{1}{g(x)}\right)+\tan ^{-1}(h(x))+c \\
\Rightarrow & \int x^{-3} \log \left(1+x^4\right) d x \\
= & \log \left(1+x^4\right) \cdot \frac{x^{-2}}{-2}+\frac{1}{2} \int\left(\frac{4 x^3}{1+x^4} \cdot x^{-2}\right) d x \\
= & -\frac{1}{2 x^2} \log \left(1+x^4\right)+\int \frac{2 x}{1+x^4} d x
\end{aligned}$
Let $x^2=t \Rightarrow 2 x=\mathrm{dt}$
$=-\frac{1}{2 x^2} \log \left(1+x^4\right)+\int \frac{d t}{1+t^2}$
$\begin{aligned} & =-\frac{1}{2 x^2} \log \left(1+x^4\right)+\tan ^{-1}\left(x^2\right)+c \\ & \text { So } f(x)=\frac{1}{2 x^2}, g(x)=1+x^4, h(x)=x^2 \\ & =x^2\left(\frac{1}{2 x^2}+\frac{x^2}{2}\right)=\frac{1+x^4}{2}=\frac{g(x)}{2}\end{aligned}$
Hence, the correct answer is option: 'B'.