- #1
karush
Gold Member
MHB
- 3,269
- 5
$\tiny\text{LCC 206 {r9} IBP}$
$I=\int \frac{x}{2\sqrt{x+2}}\,dx
=\frac{\left(x-4\right)\sqrt{x+3}}{3}$
Rewrite as
$I=\int \frac{x}{2}\frac{1}{\sqrt{x+2}}\, dx $
Let
$\displaystyle
u=\frac{x}{2} \ \ \ dv=\frac{1}{\sqrt{x+2}} \, dx \\
du=\frac{1}{2} \ \ \ v=2\sqrt{x+2} \\ $
Then
$x\sqrt{x+2}-\int \sqrt{x+2} \ dx$
This doesn't look it's heading toward the answer
$\tiny\text
{from Surf the Nations math study group}$
$I=\int \frac{x}{2\sqrt{x+2}}\,dx
=\frac{\left(x-4\right)\sqrt{x+3}}{3}$
Rewrite as
$I=\int \frac{x}{2}\frac{1}{\sqrt{x+2}}\, dx $
Let
$\displaystyle
u=\frac{x}{2} \ \ \ dv=\frac{1}{\sqrt{x+2}} \, dx \\
du=\frac{1}{2} \ \ \ v=2\sqrt{x+2} \\ $
Then
$x\sqrt{x+2}-\int \sqrt{x+2} \ dx$
This doesn't look it's heading toward the answer
$\tiny\text
{from Surf the Nations math study group}$
Last edited: