The integral dxx^2(x^4+1)^3/4 equals:

Question

The integral $\displaystyle\int \dfrac{dx}{x^2(x^4+1)^{3/4}}$ equals:

Accepted Answer

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

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

Let ${x^{ - 4}} = y$

$$ \Rightarrow  - 4{x^{ - 3}}\,dx = dy$$

$$ \Rightarrow dx = {{ - 1} \over 4}{x^3}dy$$

$$\therefore$$ $$I = {{ - 1} \over 4}\int {{{{x^3}dy} \over {{x^3}{{\left( {1 + y} \right)}^{3/4}}}}} $$

$$ = {{ - 1} \over 4}\int {{{dy} \over {{{\left( {1 + y} \right)}^{3/4}}}}} $$

$$ = {{ - 1} \over 4} \times 4{\left( {1 + y} \right)^{1/4}}$$

$$ =  - 1{\left( {1 + {x^{ - 4}}} \right)^{1/4}} + C$$

$$ =  - {\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{1/4}} + C$$

Solutions · 1