It has been about 2 years since i last did calculus but im trying to get back into it so im ready for college / dont kill myself because of boredom(adsbygoogle = window.adsbygoogle || []).push({});

I am having difficulty finding integrals of the form

[tex] \int{\frac{x^2}{x+a}dx} [/tex]

This integral inparticular is:

[tex]\int{{\frac{x^2}{2x+2}dx}[/tex]

I couldnt find a [itex]u[/itex] or a [itex] du [/itex] that would work

I then tried to decompose the fraction to [itex] \frac{x^2}{2x+2} = \frac{A}{2} + \frac{B}{x+1}[/itex] but could't solve for A, which makes sense, so that wont work.

Next i tied integration by parts

[tex] \int{vdu} = uv - \int{udv} [/tex]

where i used [itex] u=.5 ln|2x+2|[/itex],[itex]du=(2x+2)^{-1}[/itex],[tex]v=x^2[/itex],[itex]dv=2x[/itex].

then

[tex]\int{\frac{x^2}{2x+2}dx} = .5x^2 ln |2x+2| - \int{ln |2x+2|xdx} [/tex]

for the second integral I have:

[itex] u=.5x^2[/itex],[itex]du=xdx[/itex],[tex]v=ln |2x+2| [/itex],[itex]dv=2(2x+2)^{-1}[/itex].

where

[tex]\int{ x ln|2x+2| dx} = .5x^2 ln |2x+2| - \int{\frac{x^2}{2x+2}dx} [/tex]

and finally both sides cancel out:

[tex]\int{\frac{x^2}{2x+2}dx} = .5x^2 ln |2x+2| - .5x^2 ln |2x+2| +\int{\frac{x^2}{2x+2}dx} [/tex]

[tex] \int{\frac{x^2}{2x+2}dx} = \int{\frac{x^2}{2x+2}dx} [/tex]

Ive checked the trig rules as well as the inverse trig rules and no other rules match. Does an integral for this function not exsist or am I missing something?

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# Homework Help: Remembering how to integrate

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