1
$\begingroup$

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

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

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

$du=dx$

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

How can I continue from here?

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.