How to integrate $\int\frac{x}{(x^2-2x+5)^2} \, dx$?

Question

$$\int\frac{x}{(x^2-2x+5)^2} \, dx$$ I tried to complete the square of the bottom like this $\int\frac{x}{((x-1)^2+4)^2} \, dx$ but I'm still not sure what to do.

Any help would be appreciated.

Thanks

Answer 1

The below solution is obtained from WolframAlpha.

Answer 2

The standard first calculus course substitution is $x-1=2\tan t$. Then $dx=2\sec^2 t$, and $(x-1)^2+4=4\sec^2 t$. So our integral is equal to $$\int \frac{(1+2\tan t)(2\sec^2 t)}{16\sec^4 t}\,dt.$$

Separate into two parts in the natural way. The integral $$\int \frac{2}{16}\frac{1}{\sec^2 t}\, dt$$ is familiar, we need to integrate $\frac{2}{16}\cos^2 t$.

The remaining integral is $$\int \frac{4}{16}\frac{\tan t}{\sec^2 t}\,dt$$ is also done by switching to the more familiar sines and cosines. We end up needing $$\int\frac{4}{16}\sin t\cos t\,dt.$$ This one is immediate: let $u=\sin t$.

Remark: It may be worthwhile at the very beginning to write the top as $(x-1)+1$. Then note that $x-1$ is almost the derivative of
$x^2-2x+5$, making the substitution $u=x^2-2x+5$ natural. So that's an easy integral.

Now all we need to do is to find the integral of $\dfrac{1}{(x^2-2x+5)^2}$.

Complete the square and make the substitution $x-1=2\tan t$ like in the main answer.

Answer 3

A good strategy, now that you've completed the square is to also "complete the differential":

$$\int\frac{x}{((x-1)^2+4)^2} \, dx=$$

$$\frac 1 2\int\frac{2(x-1)}{((x-1)^2+4)^2} \, dx+\int\frac{1}{((x-1)^2+4)^2} \, dx=$$

$$\frac 1 2\int\frac{du}{u^2} +\int\frac{1}{(u^2+4)^2} \, du=$$

$$-\frac 1 2\frac{1}{u} +\int\frac{1}{(u^2+4)^2} \, du=$$

Now, to solve $$\int\frac{1}{(u^2+4)^2} \, du$$

take $2\tan \rho =u$

It will follow

$$\int\frac{1}{(u^2+4)^2} \, du$$

$$\int\frac{2\sec^2\rho }{16\sec^4 \rho } \,d\rho$$

$$\int\frac{1}{8\sec^2 \rho } \,d\rho=\frac 1 8\int{\cos^2\rho} \,d\rho$$

Now use the fact that $\dfrac{1+\cos(2x)}{2}=\cos^2 x$ to finish it off.