# Integration by Parts

• Feb 2nd 2011, 09:48 AM
quantoembryo
Integration by Parts
$\displaystyle \int{x^2tan^{-1}xdx}$

$\displaystyle \int{x^2tan^{-1}xdx} = \frac{x^3}{3}tan^{-1}x-\frac{1}{3}\int \frac {x^3}{1+x^2}dx$

$\displaystyle let {}u=1+x^2, \frac{du}{2}=xdx$

$\displaystyle \frac{x^3}{3}tan^{-1}x- \frac{1}{6}\int (1-1/u)$

$\displaystyle \frac{x^3}{3}tan^{-1}x-(x^2/6)-(1/6)+(1/6)ln(x^2+1)$

In the answer, there is no -1/6. Help?
• Feb 2nd 2011, 09:52 AM
TheEmptySet
Quote:

Originally Posted by quantoembryo
$\displaystyle \int{x^2tan^{-1}xdx}$

$\displaystyle \int{x^2tan^{-1}xdx} = \frac{x^3}{3}tan^{-1}x-\frac{1}{3}\int \frac {x^3}{1+x^2}dx$

$\displaystyle let {}u=1+x^2, \frac{du}{2}=xdx$

$\displaystyle \frac{x^3}{3}tan^{-1}x- \frac{1}{6}\int (1-1/u)$

$\displaystyle \frac{x^3}{3}tan^{-1}x-(x^2/6)-(1/6)+(1/6)ln(x^2+1)$

In the answer, there is no -1/6. Help?

Since you have an indefinate itegral you need an arbitarty constnat $\displaystyle +C$ the $\displaystyle -\frac{1}{6}$ can be combined into the constant.
• Feb 2nd 2011, 09:54 AM
quantoembryo
So would you always get rid of terms without a variable?
• Feb 2nd 2011, 10:00 AM
TheEmptySet
Quote:

Originally Posted by quantoembryo
So would you always get rid of terms without a variable?

Remember that anti derivatives are only unique up to an arbitrary constant.

You can check that you have a correct answer by taking the derivative.

Notice that if you take the derivative of$\displaystyle C$ or $\displaystyle C-\frac{1}{6}$ both derivatives are zero. Also notice that if you were given limits of integration you would end up with $\displaystyle \displaystyle -\frac{1}{6}-\left( -\frac{1}{6}\right)=0$

In my opinion I always combine constants in my $\displaystyle C$
• Feb 2nd 2011, 10:02 AM
TheEmptySet
Also I just noticed I don't see where you got you $\displaystyle \frac{1}{6}$. I don't think it should be there. Double check you integration done after integrating by parts
• Feb 2nd 2011, 11:38 AM
tom@ballooncalculus
Possibly the OP simplified

$\displaystyle \displaystyle{\frac{x^3}{1 + x^2}}$

to

$\displaystyle \displaystyle{1 - \frac{x}{1 + x^2}}$

which should have been

$\displaystyle \displaystyle{\frac{x(1 + x^2) - x}{1 + x^2}}\ =\ \displaystyle{x - \frac{x}{1 + x^2}}$

[Edit: no, that doesn't explain it.]

Anyway, just in case a picture helps...

http://www.ballooncalculus.org/draw/parts/ten.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/prod.png

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,

http://www.ballooncalculus.org/asy/maps/parts.png

... is lazy integration by parts, doing without u and v.

Further edit: (See below.) Yes, false alarm. Substituting where they did, the OP integrated kind of like this:

http://www.ballooncalculus.org/draw/parts/tena.png

... so the sixth was just an extra constant as The Empty Set at first thought. Pardon the fuss. (Crying)

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
• Feb 2nd 2011, 11:58 AM
Quote:

Originally Posted by quantoembryo
$\displaystyle \int{x^2tan^{-1}xdx}$

$\displaystyle \int{x^2tan^{-1}xdx} = \frac{x^3}{3}tan^{-1}x-\frac{1}{3}\int \frac {x^3}{1+x^2}dx$

$\displaystyle let {}u=1+x^2, \frac{du}{2}=xdx$

$\displaystyle \frac{x^3}{3}tan^{-1}x- \frac{1}{6}\int{(1-1/u)}du$

$\displaystyle \frac{x^3}{3}tan^{-1}x-(x^2/6)-(1/6)+(1/6)ln(x^2+1)+$C

In the answer, there is no -1/6. Help?

$\displaystyle \displaystyle\frac{2x^3tan^{-1}x-x^2+ln\left[1+x^2\right]}{6}+\left(C-\frac{1}{6}\right)$