1. ## integration by parts

use integration by parts to find each of the following integrals:

$\displaystyle \int tan^{-1} \left(\frac{x}{2}\right) dx$

$\displaystyle \int x^2 e^{2x} dx$

$\displaystyle \int log_3(2x+1) dx$

thanks : )

2. Originally Posted by qzno
use integration by parts to find each of the following integrals:

$\displaystyle \int tan^{-1} \left(\frac{x}{2}\right) dx$

$\displaystyle u = \tan^{-1}\left(\frac{x}{2}\right)$

$\displaystyle du = \frac{2}{4+x^2} \, dx$

$\displaystyle dv = dx$

$\displaystyle v = x$

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$\displaystyle \int x^2 e^{2x} dx$

$\displaystyle u = x^2$

$\displaystyle du = 2x \, dx$

$\displaystyle dv = e^{2x} \, dx$

$\displaystyle v = \frac{1}{2}e^{2x}$

btw ... this one is easier if you use tabular integration.

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$\displaystyle \int log_3(2x+1) dx$

change base ...

$\displaystyle \frac{1}{\ln{3}} \int \ln(2x+1) dx$

$\displaystyle u = \ln(2x+1)$

$\displaystyle du = \frac{2}{2x+1} \, dx$

$\displaystyle dv = dx$

$\displaystyle v = x$
go for it.

3. thank you : )

4. $\displaystyle xtan^{-1}\left(\frac{x}{2}\right) - \frac{x^2}{4} + ln x^2 + C$

5. i got something different:

$\displaystyle xtan^{-1}\left(\frac{x}{2}\right) - \frac{1}{4} + ln x^2 + C$

6. $\displaystyle \int \tan^{-1}\left(\frac{x}{2}\right) \, dx$

$\displaystyle u = \tan^{-1}\left(\frac{x}{2}\right)$

$\displaystyle du = \frac{2}{4 + x^2} \, dx$

$\displaystyle dv = dx$

$\displaystyle v = x$

$\displaystyle \int \tan^{-1}\left(\frac{x}{2}\right) \, dx = x \cdot \tan^{-1}\left(\frac{x}{2}\right) - \int \frac{2x}{4 + x^2} \, dx$

$\displaystyle \int \tan^{-1}\left(\frac{x}{2}\right) \, dx = x \cdot \tan^{-1}\left(\frac{x}{2}\right) - \ln(4 + x^2) + C$

7. are the other two answers:

$\displaystyle \frac{1}{2}e^{2x} \left(x^2 - x + \frac{1}{2}\right) + C$

and

$\displaystyle \frac{1}{ln3}\left(xln(2x+1) - x - x^2\right) + C$

???

8. how does one check an antiderivative?

9. take the derivative of the answer.