Indefinite Integration Question 1
Question 1 - 2024 (27 Jan Shift 2)
The integral $\int \frac{\left(x^{8}-x^{2}\right) d x}{\left(x^{12}+3 x^{6}+1\right) \tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)}$ equal to :
(1) $\log _e\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)^{1 / 3}+C$
(2) $\log _e\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)^{1 / 2}+C$
(3) $\log _e\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)+C$
(4) $\log _e\left(\left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|\right)^{3}+C$
Show Answer
Answer (1)
Solution
$I=\int \frac{x^{8}-x^{2}}{\left(x^{12}+3 x^{6}+1\right) \tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)} d x$
Let $\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)=t$
$\Rightarrow \frac{1}{1+\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} \cdot\left(3 x^{2}-\frac{3}{x^{4}}\right) d x=d t$
$\Rightarrow \frac{x^{6}}{x^{12}+3 x^{6}+1} \cdot \frac{3 x^{6}-3}{x^{4}} d x=d t$
$I=\frac{1}{3} \int \frac{d t}{t}=\frac{1}{3} \ln |t|+C$
$I=\frac{1}{3} \ln \left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|+C$
$I=\ln \left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|^{1 / 3}+C$
Hence option (1) is correct
