STD 12 - 7.1 inderfinite integral — JEE कक्षा 11 साइन्स — Question

Taking $x^{4}$ common from denominator

$=\int \frac{1 d x}{x^{2}\left(x^{4}\right)^{\frac{3}{4}}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$

$=\int \frac{d x}{x^{2}\left(x^{3}\right)\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$

$=\int \frac{d t}{x^{5}\left(1+\frac{1}{x^{4}}\right)^{\frac{3}{4}}}$

Let $t = 1 + \frac{4}{{{x^4}}}$

$\frac{{dt}}{{dx}} =  - \frac{4}{{{x^5}}}$

$-\frac{d t}{4}=\frac{d x}{x^{5}}$

Substituting value of $x$ and $d x$

$ = \frac{{ - 1}}{4}\int {\frac{{dt}}{{{t^{\frac{3}{4}}}}}} $

${=\frac{-1}{4} \int t^{\frac{-3}{4}} d t} $

${=\frac{-1}{4}\left[\frac{t^{\frac{-3}{4}}+1}{\frac{-3}{4}+1}\right]+C} $

${=\frac{-1}{4}\left(\frac{t^{\frac{1}{4}}}{\frac{1}{4}}\right)+C}$

$=-t^{\frac{1}{4}}+c$

$=-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c$