Question

Integral of x^2 dx/x^2 + 9
Use partial fraction decomposition, and long division if needed.

Answer 1

I will post an example

Answer 2

For example:

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{x^2}{x^2+4}~dx}\)

You will perform this trig-sub:

\(\large\color{black}{\displaystyle x=\sqrt{4}\tan\theta=2\tan\theta}\)
\(\large\color{black}{\displaystyle dx/d\theta =2\sec^2\theta\quad\Longrightarrow\quad dx=2\sec^2\theta{~}d\theta }\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{\left(2\tan\theta\right)^2}{\left(2\tan\theta\right)^2+4}~\left(2\sec^2\theta{~}d\theta\right)}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{4\tan^2\theta}{4\tan^2\theta+4}~\left(2\sec^2\theta{~}d\theta\right)}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{8\tan^2\theta\sec^2\theta}{4\tan^2\theta+4}~{~}d\theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{8\tan^2\theta\sec^2\theta}{4(\tan^2\theta+1)}~{~}d\theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{8\tan^2\theta\sec^2\theta}{4(\sec^2\theta)}~{~}d\theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}2\tan^2\theta{~}d\theta}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}2\sec^2\theta+2{~}d\theta}\)

\(\large\color{black}{\displaystyle 2\tan\theta+2\theta+C}\)

Now we will rearrange \(\theta\) in terms of x:

\(\large\color{black}{\displaystyle x=2\tan\theta}\)

\(\large\color{black}{\displaystyle \frac{x}{2}=\tan\theta}\)

\(\large\color{black}{\displaystyle \theta=\arctan\left(\frac{x}{2}\right)}\)

Now do the substitution:

\(\large\color{black}{\displaystyle 2\tan\left(\arctan\left(\frac{x}{2}\right)\right)+2\arctan\left(\frac{x}{2}\right)+C}\)

\(\large\color{black}{\displaystyle 2\cdot\frac{x}{2}+2\arctan\left(\frac{x}{2}\right)+C}\)

\(\large\color{black}{\displaystyle x+2\arctan\left(\frac{x}{2}\right)+C}\)

Answer 3

I just used a 4 instead of a 9, but I hope that this helps.

Answer 4

Trig substitution, yeah... I'm just not sure if I'm supposed to solve the problem with just the long division/partial fractions. I guess not, if you're not seeing a way to do it that way.

Answer 5

I don't think you can do partial fractions on this because the bottom can not be factored.
Trig sub will be needed

Answer 6

\(\rm x^2+4\) is not factorable (well, not into reals)

Answer 7

Right. Thanks!

Answer 8

You Welcome !

Answer 9

this is wrong

Answer 10

Are you sure that this is wrong?
I really doubt that there is any error in my example, although anyone can make a mistake. If you think that I really made a mistake, then please pint out where and how.

Answer 11

nooooooooooooooo

Answer 12

you're fine

Answer 13

Then what is wrong? To do what I did is a violation of the site's policy?

Answer 14

wat o.O

Answer 15

my love for you is wrong ;-; k bye

Answer 16

\(\Large\color{black}{\displaystyle\lim_{a \rightarrow ~\infty}\left( x{\rm D} \right)^a}\)

Answer 17

awh :') ofc our love is beyond infinity <3