I have trouble with both problems:
1. $\displaystyle \int ln(x+3)dx$
2. $\displaystyle \int (x-1)sinxdx$
Dear littlesohi,
1)$\displaystyle \int{ln(x+3)dx}=(x+3)ln(x+3)-(x+3)$
2)$\displaystyle \int{(x-1)Sinxdx}$
$\displaystyle \int{xSinxdx}-\int{Sinxdx}$
$\displaystyle -xCosx+\int{Cosxdx}+Cosx+C\mbox{ where C is an arbitary constant.}$
$\displaystyle -xCosx+Sinx+Cosx+C$
Hope this helps.
Dear littlesohi,
I didn't use integration by parts in the first problem. But you could use the integration by parts method as well.
$\displaystyle \int{ln(x+3)dx}$
$\displaystyle \int{\frac{dx}{dx}\ln(x+3)dx}$
$\displaystyle xln(x+3)-\int{\frac{x}{x+3}dx}$
I think you could do it from here.
$\displaystyle u=\ln(x+3) ~\implies~ du=\frac{dx}{x+3}$
$\displaystyle dv=dx ~\implies~ v=x$
Giving the result:
$\displaystyle uv - \int v \; du ~=~ x \ln(x+3) - \int \frac{x}{x+3} \; dx$
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.
Integration by parts is derived from the
Product Rule.
(not the Chain Rule as i originally wrote for unknown reasons!!)
$\displaystyle \frac{d}{dx}uv=u\frac{dv}{dx}+v\frac{du}{dx}$
$\displaystyle v\frac{du}{dx}=\frac{d}{dx}uv-u\frac{dv}{dx}$
$\displaystyle \int{v\frac{du}{dx}dx}=\int{\frac{d}{dx}(uv)dx}-\int{u\frac{dv}{dx}dx}$
$\displaystyle \int{udv}=\int{d(uv)}-\int{vdu}$
$\displaystyle \int{vdu}=uv-\int{udv}$
or
$\displaystyle \int{udv}=uv-\int{vdu}$
$\displaystyle \int{ln(x+3)dx}=\int{udv}$
$\displaystyle u=ln(x+3),\ dv=dx,\ v=x$
$\displaystyle u=ln(x+3),\ \frac{du}{dx}=[\frac{1}{x+3}\ \frac{d}{dx}(x+3)]=\frac{1}{x+3}$
$\displaystyle du=\frac{1}{x+3}dx$
$\displaystyle uv-\int{vdu}=xln(x+3)-\int{\frac{x}{x+3}}dx$