Q75 / 120 · VITEEE 2016 · Mathematics
$\int \frac{1}{x^{2}\left(x^{4}+1\right)^{3 / 4}} d x$ is equal to
- A$\left(1+\frac{1}{x^{4}}\right)^{1 / 4}+C$
- B$\left(x^{4}+1\right)^{1 / 4}+C$
- C$\left(1-\frac{1}{x^{4}}\right)^{1 / 4}+C$
- Correct answerD$-\left(1+\frac{1}{x^{4}}\right)^{1 / 4}+C$
Solution
$\int \frac{d x}{x^{2}\left(x^{4}+1\right)^{3 / 4}}=\int \frac{d x}{x^{5}\left(1+\frac{1}{x^{4}}\right)^{3 / 4}}$
Put $1+\frac{1}{x^{4}}=t \Rightarrow-\frac{4}{x^{5}} d x=d t$
So, integral is
$$
\mathrm{I}=-\frac{1}{4} \int \frac{\mathrm{dt}}{\mathrm{t}^{3 / 4}}=-\mathrm{t}^{\frac{1}{4}}+\mathrm{c}=-\left(1+\frac{1}{\mathrm{x}^{4}}\right)^{1 / 4}+\mathrm{c}
$$
Hence, the correct answer is option: 'D'.