Question

Prove that $\int \frac{x^{24}}{x^{10}+1}dx=[\frac{t^3}{3}-t+{\tan }^{-1}t]wheret=x^5$.

Answer 1

Let $I=\int \frac{x^{24}}{x^{10}+1}dx$

$=\int \frac{x^{20}.x^{4}dx}{{\left(x^5\right)}^2+1}$
Put $x^5=t$

$\Rightarrow 5x^{4}dx=dt$
$\Rightarrow I=\frac{1}{5}\int \frac{t^{4}dt}{t^2+1}$

$=\frac{1}{5}\int \frac{t^4-1+1}{t^2+1}dt$

$=\frac{1}{5}\int (t^2-1+\frac{1}{t^2+1})dt$
$=\frac{1}{5}[\frac{t^3}{3}-t+{\tan }^{-1}t]$ where $t=x^5$