Answer
$$\int\frac{x^2}{\sqrt{x^2-4x+5}}dx=-\frac{1}{2}\ln(x-2+\sqrt{(x-2)^2+1})+\frac{(x-2)\sqrt{(x-2)^2+1}}{2}+4\sqrt{(x-2)^2+1}+4\sinh^{-1}(x-2)+C$$
Work Step by Step
$$A=\int\frac{x^2}{\sqrt{x^2-4x+5}}dx$$ $$A=\int\frac{x^2}{\sqrt{(x-2)^2+1}}dx$$
We set $u=x-2$, which means $$du=dx$$
Also, $x=u+2$ and $x^2=u^2+4u+4$
Therefore, $$A=\int\frac{u^2+4u+4}{\sqrt{u^2+1}}du$$ $$A=\int\frac{u^2}{\sqrt{u^2+1}}du+4\int\frac{u}{\sqrt{u^2+1}}du+4\int\frac{du}{\sqrt{u^2+1}}$$
- For the first integral, use Formula 39:
$$\int \frac{x^2dx}{\sqrt{a^2+x^2}}=-\frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})+\frac{x\sqrt{a^2+x^2}}{2}+C$$
for $a=1$.
- For the second integral, set $a=\sqrt{u^2+1}$, so $da=\frac{2u}{2\sqrt{u^2+1}}du=\frac{u}{\sqrt{u^2+1}}du$
- For the third integral, use Formula 34, which states that
$$\int \frac{dx}{\sqrt{a^2+x^2}}=\sinh^{-1}\frac{x}{a}+C$$
for $a=1$.
$$A=-\frac{1}{2}\ln(u+\sqrt{u^2+1})+\frac{u\sqrt{u^2+1}}{2}+4\int da+4\sinh^{-1}u+C$$ $$A=-\frac{1}{2}\ln(u+\sqrt{u^2+1})+\frac{u\sqrt{u^2+1}}{2}+4a+4\sinh^{-1}u+C$$ $$A=-\frac{1}{2}\ln(x-2+\sqrt{(x-2)^2+1})+\frac{(x-2)\sqrt{(x-2)^2+1}}{2}+4\sqrt{(x-2)^2+1}+4\sinh^{-1}(x-2)+C$$
