How do I solve $\int\frac{7}{\sqrt{x}(x+4)}~\mathrm{d}x$?

Question

I am trying to solve $\int\frac{7}{\sqrt{x}(x+4)}~\mathrm{d}x$. So far I have $$7\int\frac{1}{\sqrt{x}(x+4)}~\mathrm{d}x$$ $$u=\sqrt{x}$$$$\mathrm{d}u=\frac{1}{2\sqrt{x}}$$ and this is where I'm not sure of what to do. This is what I tried. $$2\mathrm{d}u=\sqrt{x}$$ I am trying to solve $\int\frac{7}{\sqrt{x}(x+4)}~\mathrm{d}x$. So far I have $$14\int\frac{1}{(x+4)}~\mathrm{d}u$$ But I haven't gotten further since I'm not sure what to do. How do I go about solving this integral? Thanks in advance for all the help!

Answer 1

Hint: $I = \displaystyle 14\int \dfrac{d(\sqrt{x})}{(\sqrt{x})^2 + 4}$

Answer 2

With your change of variable we have

$$du=\frac{dx}{2\sqrt x}\implies dx=2udu$$ so the integral becomes

$$14\int\frac{du}{u^2+4}$$ Can you take it from here?

Answer 3

By substituting $x=4y^2$ you have $dx = 8y\,dy$ and: $$ I = 7\int\frac{dx}{(x+4)\sqrt{x}}=7\int\frac{dy}{y^2+1}=7\arctan(y)+C=7\arctan\sqrt{\frac{x}{4}}+C.$$