Question

*$\int \frac{1}{x^2{(x^4+1)}^{3/4}}dx=-\frac{1}{x}{(1+x^4)}^{1/4}$. If this is true enter 1, else enter 0.

Answer 1

Let $I=\int \frac{1}{x^2{(x^4+1)}^{3/4}}dx=\int \frac{dx}{x^2.x^3.{[1+\left(1/x^4\right)]}^{3/4}}$
Put $1+\frac{1}{x^4}=t\Rightarrow -\frac{4}{x^5}dx=dt$
Therefore
$I=-\frac{1}{4}\int \frac{dt}{t^{3/4}}=-\frac{1}{4},4t^{1/4}=-t^{1/4}=-{(1+\frac{1}{x^4})}^{1/4}=-\frac{1}{x}{(1+x^4)}^{1/4}$