Question

Find the indefinite integral of 2%7D%2Bx%5E%5Cfrac%7B3%7D%7B2%7D%7D%20%5C%2C%20dx%20" id="TexFormula1" title=" \int\limits {\frac{5}{x^\frac{1}{2}+x^\frac{3}{2}} \, dx " alt=" \int\limits {\frac{5}{x^\frac{1}{2}+x^\frac{3}{2}} \, dx " align="absmiddle" class="latex-formula">
I have been able to simplify it to $\int\limits {\frac{5\sqrt{x}}{x^3+x}} \, dx$ but that is confusing,

I then did u-subsitution where $u=\sqrt{x}$ to obtain $\int\limits {\frac{5u}{u^6+u^2}} \, dx$ which simplified to $\int\limits {\frac{5}{u^5+u}} \, dx$, a much nicer looking integrand
however, I am still stuck

ples help
show all work or be reported

Answer 1

The easiest way to calculate this integral is substitution.

$\int\dfrac{5}{x^\frac{1}{2}+x^\frac{3}{2}}\,dx=5\int\dfrac{1}{x^\frac{1}{2}+(x^\frac{1}{2})^3}\,dx=5\int\dfrac{1}{\sqrt{x}+(\sqrt{x})^3}\,dx=(\star)$

Now we can substitute $u=\sqrt{x}$ and then:

$du=\dfrac{1}{2\sqrt{x}}\,dx\qquad\implies\qquad dx=2\sqrt{x}\,du=2u\,du$

So:

$(\star)=5\int\dfrac{1}{\sqrt{x}+(\sqrt{x})^3}\,dx=5\int\dfrac{2u}{u+u^3}\,du=10\int\dfrac{1}{1+u^2}\,dx=$

$=10\arctan(u)+C=\boxed{10\arctan(\sqrt{x})+C}$