Problem with integration by parts

**billb91** · February 4th 2013, 10:24 PM

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

**kspkido** · February 4th 2013, 10:58 PM

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..

**Prove It** · February 4th 2013, 11:11 PM

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

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

**hollywood** · February 5th 2013, 06:24 AM

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

**billb91** · February 5th 2013, 08:17 AM

Thanks so much for the help guys! Setting it up the way prove it did made things way, way simpler!

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