How to integrate the following integrals $\int\frac{\sqrt x\,\mathrm dx}{x^5 \sqrt{1-x^7}}$

Question

integration is given

$$\int\frac{\sqrt x\,\mathrm dx}{x^5 \sqrt{1-x^7}}$$

to simplify use $x = u^2$

$$\int \frac{2\,\mathrm du}{u^8 \sqrt{1-u^{14}}}$$

same form as in the beginning. help me to solve this integration. thanks!

Answer 1

Take $u^{14}$ common from denominator square root to get:

$$\int \frac{2u^{-15} du}{\sqrt{u^{-14}-1}}$$

Now it should be easy! Take $u^{-14} -1 = t$ to get:

$$\frac{-2}{14}\int \frac{dt}{\sqrt{t}}$$

Answer 2

Hint The form $\sqrt{1 - x^7}$ suggests trying the substitution $$v = x^{7 / 2}, \qquad dv = \tfrac{7}{2} x^{5 / 2} dx ,$$ or, once you've made the substitution to $u$, $$v = u^7, \qquad dv = 7 u^6 \,du .$$

Additional hint Either way, after some (algebraic) substitution, this yields the simpler integral $$\frac{2}{7} \int \frac{dv}{v^2 \sqrt{1 - v^2}} ,$$ which can be handled, e.g., with trigonometric substitution.