# Thread: Integration by Parts

1. ## Integration by Parts

Hi guys, need some help on these:

1. $\int x (lnx) dx$

2. $\int x(cos5x)dx$

2. Originally Posted by c_323_h
Hi guys, need some help on these:

1. $\int x (lnx) dx$
Let,
$u=\ln x$ and, $v'=x$
Thus,
$u'=1/x$ and, $v=(1/2)x^2$
Thus,
$\frac{1}{2}x^2\ln x-\int \frac{1}{2} xdx$
Thus,
$\frac{1}{2}x^2\ln x-\frac{1}{4}x^2+C$

3. Originally Posted by c_323_h

2. $\int x(cos5x)dx$
Let,
$u=x$ and $v'=\cos 5x$
Thus,
$u'=1$ and $v=\frac{1}{5}\sin 5x$
Thus,
$\frac{1}{5}x\sin 5x-\int \frac{1}{5}\sin 5xdx$
Thus,
$\frac{1}{5}x\sin 5x+\frac{1}{25}\cos 5x+C$

(I am the quickest integrator in the west!)

4. Originally Posted by ThePerfectHacker
Let,

$...-\int \frac{1}{2} xdx$
why isn't it $\frac{1}{2}x^2dx$.
Where did the $x^2$ dissappear to?

5. Originally Posted by c_323_h
why isn't it $\frac{1}{2}x^2dx$.
Where did the $x^2$ dissappear to?
$\int x^n=\frac{x^{n+1}}{n+1}, n\not =-1$
Thus,
$\int xdx=\frac{1}{2}x^2$ but then you need to multiply by $\frac{1}{2}$ thus,
$\frac{1}{2}\cdot \frac{1}{2}x^2=\frac{1}{4}x^2$