Problem with integration by parts

• Feb 4th 2013, 10:24 PM
billb91
Problem with integration by parts
Hey guys/gals,

I have a problem involving integration by parts that has me stumped. Here is the equation:

$\int {xln(4+x^2)}dx$

Problem is, I assume that the u will be ln(4+x^2) and the dv will be xdx, but when I solve for that, my equation ends up being:

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

...which may be right, but I don't know what my next step should be. Any help on this would be greatly appreciated, just having a total brainfart with this one.

Edit: only took 50 tries to get both equations formatted right... you know you're a noob when... :P
• Feb 4th 2013, 10:58 PM
kspkido
Re: Problem with integration by parts
I suggest, you divide the numberator with the denominator... using division algorithm (manual division that is to say), then after you have separated the fraction.. integrate :).. you cannot decompose the function since the numerator has a higher degree.. also.. you cannot use inverse tangent.. :)
• Feb 4th 2013, 11:11 PM
Prove It
Re: Problem with integration by parts
Quote:

Originally Posted by billb91
Hey guys/gals,

I have a problem involving integration by parts that has me stumped. Here is the equation:

$\int {xln(4+x^2)}dx$

Problem is, I assume that the u will be ln(4+x^2) and the dv will be xdx, but when I solve for that, my equation ends up being:

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

...which may be right, but I don't know what my next step should be. Any help on this would be greatly appreciated, just having a total brainfart with this one.

Edit: only took 50 tries to get both equations formatted right... you know you're a noob when... :P

This should not be done with Integration by Parts, at least not to begin with...

\displaystyle \begin{align*} \int{x\ln{\left( 4 + x^2 \right) }\,dx} &= \frac{1}{2}\int{2x\ln{\left( 4 + x^2 \right) } \, dx} \end{align*}

Now let \displaystyle \begin{align*} w = 4 + x^2 \implies dw = 2x\,dx \end{align*} and the integral becomes

\displaystyle \begin{align*} \frac{1}{2}\int{2x\ln{\left( 4 + x^2 \right)}\,dx} &= \frac{1}{2} \int{\ln{(w)}\,dw} \end{align*}

Now using integration by parts with \displaystyle \begin{align*} u = \ln{(w)} \implies du = \frac{1}{w}\,dw \end{align*} and \displaystyle \begin{align*} dv = 1\,dw \implies v = w \end{align*} we have

\displaystyle \begin{align*} \frac{1}{2}\int{\ln{(w)}\,dw} &= \frac{1}{2}\left[ w\ln{(w)} - \int{w\left( \frac{1}{w}\right) dw} \right] \\ &= \frac{1}{2} \left[ w\ln{(w)} - \int{1\,dw} \right] \\ &= \frac{1}{2} \left[ w\ln{(w)} - w \right] + C \\ &= \frac{1}{2}w \left[ \ln{(w)} - 1 \right] + C \\ &= \frac{1}{2}\left( 4 + x^2 \right) \left[ \ln{ \left( 4 + x^2 \right) } - 1 \right] + C \end{align*}
• Feb 5th 2013, 06:24 AM
hollywood
Re: Problem with integration by parts
Quote:

Originally Posted by billb91
Edit: only took 50 tries to get both equations formatted right... you know you're a noob when... :P

And thank you for giving us something that's easy to read...

One suggestion: instead of "ln", you can use "\ln". It adds a little space and un-italicizes it:

$\int {xln(4+x^2)}dx$

$\int {x\ln(4+x^2)}dx$

- Hollywood
• Feb 5th 2013, 08:17 AM
billb91
Re: Problem with integration by parts
Thanks so much for the help guys! Setting it up the way prove it did made things way, way simpler!