Question

Solve it $\int \frac{1}{x^2{(x^4+1)}^{3/4}}dx$

Answer 1

$I=\int \frac{dx}{x^2{(x^4+1)}^{\frac{3}{4}}}=\int \frac{dx}{x^2.x^{4\times \frac{3}{4}}(1+\frac{1}{x^4})}I=\int \frac{dx}{x^2.x^3(1+\frac{1}{x^4})}=\int \frac{dx}{x^5(1+\frac{1}{x^4})}$

Let $\frac{1}{x^4}=t\Rightarrow \frac{-4}{x^5}dx=dt\Rightarrow dx=\frac{-x^5}{4}dt$

$I=\int \frac{-x^{5}dt}{4\times x^5(1+t)}=\frac{-1}{4}\int \frac{dt}{(1+t)}=\frac{-1}{4}\ln |1+t|+CI=\frac{-1}{4}\ln |1+\frac{1}{x^4}|+C$