Can anyone give me a step-by-step process of how to answer this question?
Use the given substitution: $u^2 = x^2+4$ to find: $$\int\frac{\sqrt{x^2+4}}{x}\:\mathrm{d}x $$
The answer is: $$\sqrt{(x^2+4)} + \ln \left|\frac{\sqrt{(x^2+4)}-2} {{\sqrt{(x^2+4)}+2}}\right| + C$$
I've been trying for the past ~2 hours to try and get the above solution; no luck yet. Any help would be really appreciated, thanks!
First we note that if we have the substitution $u^{2}=x^{2}+4$, then:
$$2u\,\mathrm{d}u=2x\,\mathrm{d}x$$
Therefore your integral becomes:
$$\int\frac{\sqrt{x^{2}+4}}{x}\:\mathrm{d}x=\frac{1}{2}\int\frac{\sqrt{x^{2}+4}}{x^{2}}2x\:\mathrm{d}x=\frac{1}{2}\int\frac{\sqrt{u^{2}}}{u^{2}-4}2u\:\mathrm{d}u$$
Simplifying, we get:
$$\int\frac{u^{2}}{u^{2}-4}\:\mathrm{d}u$$
We now need to use partial fractions to yield:
$$\frac{u^{2}}{u^{2}-4}\equiv 1+\frac{1}{u-2}-\frac{1}{u+2}$$
Therefore:
$$\int\frac{\sqrt{x^{2}+4}}{x}\:\mathrm{d}x=\int\left(1+\frac{1}{u-2}-\frac{1}{u+2}\right)\:\mathrm{d}u=u+\ln|u-2|-\ln|u+2|+C$$
Substituting back, we get:
$$\int\frac{\sqrt{x^{2}+4}}{x}\:\mathrm{d}x=\sqrt{x^{2}+4}+\ln|\sqrt{x^{2}+4}-2|-\ln|\sqrt{x^{2}+4}+2|+C$$
Using the logarithm identity: $\log(a)-\log(b)=\log\left(\frac{a}{b}\right)$, you get:
$$\int\frac{\sqrt{x^{2}+4}}{x}\:\mathrm{d}x=\sqrt{x^{2}+4}+\ln\left|\frac{\sqrt{x^{2}+4}-2}{\sqrt{x^{2}+4}+2}\right|+C$$
Using the substitution $u^2=x^2+4$, note that $$x=\sqrt{u^2-4}$$ $$(2u)du=(2x)dx\Rightarrow dx=\frac{u}{x}du=\frac{u}{\sqrt{u^2-4}}du$$ So the integral becomes $$\int \frac{u}{\sqrt{u^2-4}}\frac{u}{\sqrt{u^2-4}}du=\int \frac{u^2}{u^2-4}du=\int du+\int\frac{4}{(u+2)(u-2)}du\\=u+\int\frac{4}{(u+2)(u-2)}du$$
For the second integral, use partial fractions:- $$\int\frac{4}{(u+2)(u-2)}du=\int\frac{1}{u-2}du-\int\frac{1}{u+2}du$$
Can you take it from here?
Ok, let $u^2 = x^2 + 4 \implies 2 u du= 2x dx \implies u du = x dx$. Hence
$$ \int \frac{ \sqrt{x^2 + 4}}{x }dx = \int \frac{u}{x} \frac{ u du}{x} = \int \frac{ u^2 du}{u^2 - 4}= \int \frac{ u^2 + 4 - 4}{u^2 - 4} du = \int du + 4 \int \frac{du}{u^2-4}$$
The second integral is easily solved using partial fractions as you know: remember $u^2 - 4 = (u-2)(u+2) $
