1. ## Integration by Parts!

I have trouble with both problems:

1. $\int ln(x+3)dx$

2. $\int (x-1)sinxdx$

2. Dear littlesohi,

1) $\int{ln(x+3)dx}=(x+3)ln(x+3)-(x+3)$

2) $\int{(x-1)Sinxdx}$

$\int{xSinxdx}-\int{Sinxdx}$

$-xCosx+\int{Cosxdx}+Cosx+C\mbox{ where C is an arbitary constant.}$

$-xCosx+Sinx+Cosx+C$

Hope this helps.

3. for the one what is the u? dv?

4. Dear littlesohi,

I didn't use integration by parts in the first problem. But you could use the integration by parts method as well.

$\int{ln(x+3)dx}$

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

$xln(x+3)-\int{\frac{x}{x+3}dx}$

I think you could do it from here.

5. Originally Posted by littlesohi
for the one what is the u? dv?
$u=\ln(x+3) ~\implies~ du=\frac{dx}{x+3}$

$dv=dx ~\implies~ v=x$

Giving the result:

$uv - \int v \; du ~=~ x \ln(x+3) - \int \frac{x}{x+3} \; dx$

This is what Sudharaka got as well, but I think this is the way you are more used to seeing it done. You would need to use a substitution to finish the calculation.

6. Integration by parts is derived from the
Product Rule.

(not the Chain Rule as i originally wrote for unknown reasons!!)

$\frac{d}{dx}uv=u\frac{dv}{dx}+v\frac{du}{dx}$

$v\frac{du}{dx}=\frac{d}{dx}uv-u\frac{dv}{dx}$

$\int{v\frac{du}{dx}dx}=\int{\frac{d}{dx}(uv)dx}-\int{u\frac{dv}{dx}dx}$

$\int{udv}=\int{d(uv)}-\int{vdu}$

$\int{vdu}=uv-\int{udv}$

or

$\int{udv}=uv-\int{vdu}$

$\int{ln(x+3)dx}=\int{udv}$

$u=ln(x+3),\ dv=dx,\ v=x$

$u=ln(x+3),\ \frac{du}{dx}=[\frac{1}{x+3}\ \frac{d}{dx}(x+3)]=\frac{1}{x+3}$

$du=\frac{1}{x+3}dx$

$uv-\int{vdu}=xln(x+3)-\int{\frac{x}{x+3}}dx$

7. Originally Posted by Archie Meade
Integration by parts is derived from the
Chain Rule.

$\frac{d}{dx}uv=u\frac{dv}{dx}+v\frac{du}{dx}$
It's a pity but that is the product rule.

CB

8. Yes! it certainly is, Captain!
That's my greatest typo yet!!