How to integrate partial exponentials?

Question

$$\int \frac{dx}{\sqrt[3]{x^{5/2}(x+1)^{7/2}}}$$

I have tried taking the whole thing to the power 1/3 and then multiplying it inside the bracket, but I am unable to reach anywhere. How to proceed with such questions?

Answer 1

Raise both terms in the denominator to the $1/3$ power, then you have an integral of the form $\int {1 \over f(x) g(x)}\ dx$. Continue using integration by parts, where the functions are $u'(x) = 1/f(x)$ and $v(x) = 1/g(x)$ of the form $\int u'(x) v(x)\ dx$ where the $u'(x)$ is solved by an integral $u(x) v(x) - \int v'(x) u(x)\ dx$.

So here: $u'(x) = 1/x^{5/6} = x^{-5/6}$ so $u(x) = 6 x^{1/6}$ and $v(x) = 1/(1+x)^{7/6}$. Next use the general formula for integration by parts:

The first term is $u(x)v(x) = 6 x^{1/6}/(x+1)^{7/2}$. The second term is $\int u(x) v'(x)\ dx = \int 6 x^{1/6} v'(x)\ dx$, where $v'(x) = -\frac{7}{6 (x+1)^{13/6}}$. Then substitute into this last integral.

$$\frac{6 x (x+1)}{\sqrt[3]{x^{5/2} (x+1)^{7/2}}}$$

Answer 2

The integrand is $x^{-5/6}(x+1)^{-7/6}$. After shamelessly cheating with WA, we investigate a solution of the form

$$x^\alpha(x+1)^\beta$$ that gives the derivative

$$(\alpha x+\alpha+\beta x)x^{\alpha-1}(x+1)^{\beta-1}.$$

We can try to make the first factor proportional to $1$, to $x$ or to $x+1$. Then with $\alpha+\beta=0$, we have $\alpha=\frac16$ and $\beta=-\frac16$ that fit by magic.