# intergate $\int \frac{x}{(x^2-3x+17)^2}\ dx$

$$\int \frac{x}{(x^2-3x+17)^2}\ dx$$

$$\int \frac{x}{(x^2-3x+17)^2}\ dx=\int \frac{x}{((x-\frac{3}{2})^2+\frac{59}{4})^2}\ dx$$

$u=x-\frac{3}{2}$

$du=dx$

$$\int \frac{u+\frac{3}{2}}{((u)^2+\frac{59}{4})^2}\ du$$

How can I continue from here?

## 3 Answers

Notice, $$\int \frac{x}{(x^2-3x+17)^2}\ dx$$ $$=\frac{1}{2}\int \frac{(2x-3)+3}{(x^2-3x+17)^2}\ dx$$ $$=\frac{1}{2}\left(\int \frac{(2x-3)}{(x^2-3x+17)^2}\ dx+3\int \frac{1}{(x^2-3x+17)^2}\ dx\right)$$ $$=\frac{1}{2}\left(\int \frac{d(x^2-3x+17)}{(x^2-3x+17)^2}+3\int \frac{1}{\left(\left(x-\frac{3}{2}\right)^2+\frac{59}{4}\right)^2}\ dx\right)$$ $$=\frac{1}{2}\left( -\frac{1}{x^2-3x+17}+3\int \frac{d\left(x-\frac 32\right)}{\left(\left(x-\frac{3}{2}\right)^2+\frac{59}{4}\right)^2}\right)$$ Now, the integral can be solved using reduction formula: $\int\frac{dx}{(a+x^2)^n}\ dx=\frac{1}{2a(n-1)}\left(\frac{x}{(a+x^2)^{n-1}}+(2n-3)\int \frac{1}{(a+x^2)^{n-1}}\ dx\right)$ as follows $$\int \frac{d\left(x-\frac 32\right)}{\left(\left(x-\frac{3}{2}\right)^2+\frac{59}{4}\right)^2}$$ $$=\frac{1}{2\left(\frac{59}{4}\right)(2-1)}\left(\frac{x-\frac32}{\left(x-\frac 32\right)^2+\frac{59}{4}}+(2\cdot 2-3)\int \frac{d\left(x-\frac 32\right)}{\left(x-\frac 32\right)^2+\frac{59}{4}}\right)$$

$$=\frac{2}{59}\left(\frac{x-\frac32}{x^2-3x+17}+\frac{1}{\frac{\sqrt{59}}{2}}\tan^{-1}\left(\frac{x-\frac 32}{\frac{\sqrt{59}}{2}}\right)\right)$$ $$=\frac{2}{59}\left(\frac{2x-3}{2(x^2-3x+17)}+\frac{2}{\sqrt{59}}\tan^{-1}\left(\frac{2x-3}{\sqrt{59}}\right)\right)$$ Now, substitute the above value of integral & simplify to get the answer.

One has $$\int \frac{u+\frac{3}{2}}{u^2+\frac{59}{4}} du = \int \frac{u}{u^2+\frac{59}{4}}du + \int \frac{\frac{3}{2}}{u^2+\frac{59}{4}}du.$$ The first term can be computed be setting $v = u^2+\frac{59}{4}$ and the second thanks to the $\arctan$ function.

Separate the integral into two integrals:

$\displaystyle\int\frac{u}{u^2 + \left(\sqrt{\frac{59}{4}}\right)^{2}} + \displaystyle\int\frac{\frac{3}{2}}{u^2 + \left(\sqrt{\frac{59}{4}}\right)^{2}}$

Use another $u$-substitution on the first integral and then use trig substitution on the second one.