MHB 206.08.05.44 partial fraction decomposition

[karush]

$\tiny{206.08.05.44}$
$\textsf{Use the method of partial fraction decomposition}\\$
\begin{align*}
\displaystyle
I_{44}
&=\int \frac{4x^3+6x^2+128x}{x^5+32x^3+256x}dx\\
&=4\int \frac{1}{x^2+16} \, dx
+6\int \frac{x}{(x^2+16)^2} \, dx
+64\int \frac{1}{(x^2+16)^2} \, dx
\end{align*}$\textsf{so far ? but before using u subst can we use $\ln$ on this}$

 

[Greg]

Good!

Suggestions:

Write as

$$\int\frac{4}{x^2+16}\text{ d}x$$

$$3\int\frac{2x}{(x^2+16)^2}\text{ d}x$$

$$64\int\frac{1}{(x^2+16)^2}\text{ d}x$$

First one: Differentiate $\arctan\left(\frac x4\right)$

Second one: $u=x^2+16$

Third one: $4\tan\theta=x$

 

[karush]

so then,
$$\int\frac{4}{x^2+16}\text{ d}x=\arctan\left(\frac x4\right)$$

with $u=x^2+16 \therefore du=2x \, dx$

$$3\int\frac{2x}{(x^2+16)^2}\text{ d}x=3\int\frac{1}{u^2} du = \frac{3}{u}$$

with $x=4\tan\theta \therefore dx=4 \sec^2{x} dx$

$$64\int\frac{1}{(x^2+16)^2}\text{ d}x
=64\int \frac{4 \sec^2{x}}{(16(\tan^2\left({x}\right)+1))^2} \, dx
=4^3\int\frac{4\sec^2(x)}{4^4 sec^4(x)} \, dx
=\int \cos^2(x) \ dx
=\frac{\sin(x)\cos{x}}{2}+\frac{x}{2}$$

$\textit{this is as far as I got... loan time on library pc ran out,,, but so far?}$

$\textit{integral calculator returned }$

$\dfrac{2x-3}{x^2+16}+\dfrac{3\arctan\left(\frac{x}{4}\right)}{2}+C$

$\textit{but don't see how it got there?}$

 

[Greg]

First one's ok. You've made a sign error on the second one and haven't completed the back-sub. For the third, you should be integrating with respect to theta. I'm sure this is a notation error but at any rate, get rid of it and continue with the back-sub.

 

[karush]

with $u=x^2+16 \therefore du=2x \, dx$ $3\int\frac{2x}{(x^2+16)^2}\text{ d}x=3\int\frac{1}{u^2} du = \frac{3}{u}$ back substitute $u=x^2+16$
$\frac{3}{x^2+16}$with $x=4\tan\theta \therefore dx=4 \sec^2{\theta} d\theta$$64\int\frac{1}{(x^2+16)^2}\text{ d}x
=64\int \frac{4 \sec^2{\theta}}{(16(\tan^2\left({\theta}\right)+1))^2} \, d\theta
=4^3\int\frac{4\sec^2(\theta)}
{4^4 sec^4(\theta)} \, d\theta
=\int \cos^2(\theta) \ dx
=\frac{\sin(\theta)\cos{\theta}}{2}+\frac{\theta}{2}$
back substitute $\theta=\arctan\left(\frac{x}{4}\right)$$\frac{2x}{x^2+16}$then finally, $\arctan\left(\frac{x}{4}\right)+\frac{3}{x^2+16}+\frac{2x}{x^2+16}+C$

something is missing!

 

[Greg]

Differentiate $3/u$.

What is $\frac{\sin(\theta)\cos(\theta)}{2}+\frac{\theta}{2}$ in terms of $x$?

 

[karush]

Differentiate $3/u$.

What is $\frac{\sin(\theta)\cos(\theta)}{2}+\frac{\theta}{2}$ in terms of $x$?

$\displaystyle\theta=\arctan\left(\frac{x}{4}\right)$
then
$\displaystyle\frac{\sin(\theta)\cos(\theta)}{2}+\frac{\theta}{2}=
\frac{\arctan\left(\frac{x}{4}\right)}{2}+\frac{2x} {x^2+16}$