Question

Solve $\int \frac{dx}{x^2(x^4+1{)}^{\frac{3}{4}}}=-{(1+\frac{1}{x^4})}^{\frac{1}{m}}$. Find $m$

Answer 1

Consider, $I=\int \frac{1}{x^2{(x^4+1)}^{3/4}}$

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

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

Let $1+\frac{1}{x^4}=t$

On differentiating w.r.t $x$

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

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

$=\frac{-1}{4}\frac{t^{-3/4+1}}{\frac{-3}{4}+1}+c$

$=t^{1/4}+c$

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

$\Rightarrow m=4$

Hence, the answer is $4.$