Q115 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
$\int \frac{x^5+x}{x^8+1} d x=$
- Correct answerA$\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^4-1}{\sqrt{2} x^2}\right)+c$
- B$\log \left(x^5+x^2\right)-\log \left(x^3+x\right)+\log (x+1)+c$
- C$\frac{2}{9} x^8-\frac{4}{9} x^6+\frac{1}{9} x^4-\frac{1}{3} x^2+c$
- D$\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^5-1}{\sqrt{2} x^3}\right)+c$
Solution
Let $x^2=8 \Rightarrow 2 x d x=d t$
$\begin{aligned} & I=\frac{1}{2} \int \frac{\left(t^2+1\right) d t}{\left(t^4+1\right)}=\frac{1}{2} \int \frac{\left(1+\frac{1}{t^2}\right) d t}{\left(t^2+\frac{1}{t^2}\right)} \\ & \Rightarrow L e t ~ t-\frac{1}{t}=u \Rightarrow\left(1+\frac{1}{t^2}\right) d t=d u \\ & \mathbb{I}=\frac{1}{2} \int \frac{d u}{u^2+2}=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+c \\ & =\frac{1}{2 \sqrt{2}} \tan ^{-1}\left\{\left(t-\frac{1}{t}\right) \times \frac{1}{\sqrt{2}}\right\}+c \\ & =\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^4-1}{\sqrt{2} x^2}\right)+c\end{aligned}$
Hence, the correct answer is option: 'A'.