Q111 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics
$\int \frac{y^2+\sqrt[3]{y^4}+\sqrt[6]{y^2}}{y\left(1+\sqrt[3]{y^2}\right)} d y=$
- Correct answerA$\frac{3}{4} \sqrt[3]{y^4}+3 \tan ^{-1}(\sqrt[3]{y})+C$
- B$\frac{3}{2} y^{2 / 3}+6 \tan ^{-1}\left(\sqrt[6]{y^2}\right)+C$
- C$\frac{2}{3 \sqrt[3]{y^2}}+6 \log \left(1+y^2\right)+C$
- D$\frac{3}{1+y}+\tan ^{-1}\left(\sqrt[3]{y^2}\right)+C$
Solution
Let
$\begin{aligned} I & =\int \frac{y^2+\sqrt[3]{y^4}+\sqrt[6]{y^2}}{y\left(1+\sqrt[3]{y^2}\right)} d y \\ & =\int \frac{y^2+y^{4 / 3}+y^{1 / 3}}{y\left(1+y^{2 / 3}\right)} d y \\ & =\int \frac{y^{4 / 3}\left(y^{2 / 3}+1\right)+y^{1 / 3}}{y\left(1+y^{2 / 3}\right)} d y \\ & =\int\left(y^{1 / 3}+\frac{y^{-2 / 3}}{1+y^{2 / 3}}\right) d y \\ & =\frac{y^{4 / 3}}{4 / 3}+\int \frac{y^{-2 / 3}}{1+\left(y^{1 / 3}\right)^2} d y \\ & =\frac{3}{4} \sqrt[3]{y^4}+3 \int \frac{d\left(y^{1 / 3}\right)}{1+\left(y^{1 / 3}\right)^2} \\ & =\frac{3}{4} \sqrt[3]{y^4}+3 \tan ^{-1} y^{1 / 3}+C \\ & =\frac{3}{4} \sqrt[3]{y^4}+3 \tan ^{-1} \sqrt[3]{y}+C\end{aligned}$
Hence, the correct answer is option: 'A'.