[ The integral (x^8-x^2) d x(x^12+3 x^6+1) ^-1(x^3+(1)/(x^3)) is equal to : ]

Question

$$\text { The integral } \int \frac{\left(x^8-x^2\right) \mathrm{d} x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} \text { is equal to : }$$

Accepted Answer

$$I=\int \frac{x^8-x^2}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} d x$$

$$\begin{aligned}
& \text { Let } \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)=t \
& \Rightarrow \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2} \cdot\left(3 x^2-\frac{3}{x^4}\right) d x=d t \
& \Rightarrow \frac{x^6}{x^{12}+3 x^6+1} \cdot \frac{3 x^6-3}{x^4} d x=d t \
& I=\frac{1}{3} \int \frac{d t}{t}=\frac{1}{3} \ln |t|+C \
& I=\frac{1}{3} \ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|+C \
& I=\ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|^{1 / 3}+C
\end{aligned}$$

Hence option (1) is correct.

Solutions · 1