Question

Integrate the following function $w.r.tx$
$\frac{x^3}{16x^8-25}$

Answer 1

Let $I=\int \frac{x^3}{16x^8-25}dx...\left(1\right)$

Let $u=x^4$

$\therefore du=4x^{3}dx$

$\therefore x^{3}dx=\frac{du}{4}$

or,$I=\frac{1}{4}\int \frac{du}{16u^2-25}$

or, $I=\frac{1}{4}\int \frac{du}{{\left(4u\right)}^2-{\left(5\right)}^2}$

$[\because \int \frac{dx}{x^2-a^2}=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|]+c$

or, $I=\frac{1}{4}\times \frac{1}{2\left(5\right)}log\left|\frac{4u-5}{4u+5}\right|+c$

$\therefore =\frac{1}{4}log\left|\frac{4x^4-5}{4x^4+5}\right|+c$ Ans (where $c=$ constant of integration)