Integration Question (1 Viewer)

Saturn WY15 · Apr 9, 2012

How do you do this question \int x^3/(x^2-x-3) ? Thanks, a better picture of the question is in the attachment

Saturn WY15 · Apr 9, 2012

Nvm I got it

SpiralFlex · Apr 9, 2012

$\int \frac{x^3}{x^2-x-3}\;dx$

By division,

$=\int (x+1+\frac{4x+3}{x^2-x-3})\;dx$

$=\frac{1}{2}x^2+x+\int \frac{4x+3}{x^2-x-3}\;dx$

$=\frac{1}{2}x^2+x+\int \frac{2(2x-1)}{x^2-x-3}\;dx+\int \frac{5}{x^2-x-3}\;dx$

$=\frac{1}{2}x^2+x+2\ln(x^2-x-3)+5\int \frac{1}{x^2-x-3}\;dx$

$=\frac{1}{2}x^2+x+2\ln(x^2-x-3)+5\int \frac{1}{x^2-x+\frac{1}{4}-\frac{1}{4}-3}\;dx$

$=\frac{1}{2}x^2+x+2\ln(x^2-x-3)+5\int \frac{1}{(x-\frac{1}{2})^2-\frac{13}{4}}\;dx$

$=\frac{1}{2}x^2+x+2\ln(x^2-x-3)+\frac{5\sqrt{13}}{13}\ln|\frac{x-\frac{1}{2}-\frac{\sqrt{13}}{2}}{x-\frac{1}{2}+\frac{\sqrt{13}}{2}}|+C$

$=\frac{1}{2}x^2+x+2\ln(x^2-x-3)+\frac{5\sqrt{13}}{13}\ln|\frac{2x-1-\sqrt{13}}{2x-1+\sqrt{13}}|+C$

Note: I used this for the third last line to the second.

$\int \frac{1}{x^2-a^2}\;dx=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C$

Drongoski · Apr 9, 2012

Excellent work as usual. Unable to rep you.

SpiralFlex · Apr 9, 2012

:O Hyperbolic tan.

D94 · Apr 9, 2012

SpiralFlex said:
Note: I used this for the third last line to the second.

$\int \frac{1}{x^2-a^2}\;dx=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C$

You better first clarify (ie. with MANSW) if this is a quotable expansion formula for that partial fractions integral. The standard integrals are quotable; others require the full expansion.

Saturn WY15 · Apr 9, 2012

Thanks guys, another question

note it has dx at the end using x=sin theta

Trebla · Apr 9, 2012

Do you have to use $x=\sin\theta$ ? It can easily be computed by reverse chain rule.

$\displaystyle\int x(1-x^2)^{-0.5}\,dx=-\sqrt{1-x^2}+c$

If you really must use substitution then

$x=\sin\theta \Rightarrow dx=\cos\theta\,d\theta \\\\\displaystyle\int \dfrac{x}{\sqrt{1-x^2}}\,dx\\\\=\displaystyle\int \dfrac{\sin\theta\cos\theta}{\sqrt{1-\sin^2\theta}}\,d\theta\:\:*\\\\ $(1) First assume that $\cos\theta > 0 \Rightarrow \cos\theta = \sqrt{1-\sin^2\theta}\\\\= \displaystyle\int \sin\theta\,d\theta\\\\=-\cos\theta + c\\\\=-\sqrt{1-\sin^2\theta}+c\\\\=-\sqrt{1-x^2}+c\\\\$(2) Now assume that $\cos\theta < 0 \Rightarrow \cos\theta = -\sqrt{1-\sin^2\theta}$ so for line * we get$\\\\-\displaystyle\int \sin\theta\,d\theta\\\\=\cos\theta + c\\\\=-\sqrt{1-\sin^2\theta}+c\\\\=-\sqrt{1-x^2}+c\\\\\therefore $ In both cases $\displaystyle\int \dfrac{x}{\sqrt{1-x^2}}\,dx=-\sqrt{1-x^2}+c\\\\$Note that we ignore the case where $\cos\theta = 0$ as this yields a zero denominator.$

deswa1 · Apr 9, 2012

I don't know if I read the question correctly (hard to see) but here goes:
<img src="http://latex.codecogs.com/gif.latex?\int \frac{x}{\sqrt[]{1-x^2}}=\frac{1}{2}\int \frac{2x}{\sqrt[]{1-x^2}}\\ =-\sqrt{1-x^2}+C" title="\int \frac{x}{\sqrt[]{1-x^2}}=\frac{1}{2}\int \frac{2x}{\sqrt[]{1-x^2}}\\ =-\sqrt{1-x^2}+C" />

Saturn WY15 · Apr 9, 2012

@ Deswa1 , You're are correct, how did u do it ?

Shadowdude · Apr 9, 2012

deswa1 said:
I don't know if I read the question correctly (hard to see) but here goes:
<img src="http://latex.codecogs.com/gif.latex?\int \frac{x}{\sqrt[]{1-x^2}}=\frac{1}{2}\int \frac{2x}{\sqrt[]{1-x^2}}\\ =-\sqrt{1-x^2}+C" title="\int \frac{x}{\sqrt[]{1-x^2}}=\frac{1}{2}\int \frac{2x}{\sqrt[]{1-x^2}}\\ =-\sqrt{1-x^2}+C" />

$\frac{1}{2} \int \frac{2x}{\sqrt{1-x^{2}}}\;dx$

$\textup{Let}\,u=1-x^{2}\;. \textup{So:}\, du = 2x dx$

$\frac{1}{2} \int \frac{1}{\sqrt{u}}\;du = \frac{1}{2} \int u^{-1/2} du$

And so on...

deswa1 · Apr 9, 2012

The way I do questions like this is by recognition. Notice that you have (1-x^2)^0.5 on the denominator and an x on the top. When you differentiate (1-x^2)^0.5, it will become (1/2)*(1-x^2)^-0.5*(-2x). This is very similar to the given integral (just missing the minus sign in fact). Therefore you adjust constants to make the question work.

When you do a few of tehse questions, you'll pick it up fast. If not, you can try using a substitution like u^2=1-x^2 or u=1-x^2 until you get used to doing it mentally.

Saturn WY15 · Apr 9, 2012

wait you have to use x=sin theta

Shadowdude · Apr 9, 2012

what, now i'm confused... What are we solving?

mirakon · Apr 9, 2012

Same question, different substitution

Saturn WY15 · Apr 9, 2012

don't worry i got it

Saturn WY15 · Apr 9, 2012

i just didn't understand the last step where u got cos theta and u have to change it back in to x form using the knowledge that x=sin theta

Siddy123 · Apr 10, 2012

below is a link for some great integration questions.

http://temp-share.com/show/FHKdPGm56

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