# Thread: ∫x(x+4)^(-2) dx Find the indefinite integral

1. ## ∫x(x+4)^(-2) dx Find the indefinite integral

∫x(x+4)^(-2) dx Find the indefinite integral.

I got (-x/x+4)+(x+4)+C. I was only given half credit, so I'm wondering if this drastically simplifies or something.

2. See attachment--- what did you do?

3. This is being solved in the section of my book using uv- ſ v du form. I went about trying to solve it that way. The way you solved it is not something we have learned (it is bus cal).

4. Originally Posted by jebckr
This is being solved in the section of my book using uv- ſ v du form. I went about trying to solve it that way. The way you solved it is not something we have learned (it is bus cal).
$u = x$ ... $dv = \frac{1}{(x+4)^2} \, dx$

$du = dx$ ... $v = -\frac{1}{x+4}$

$uv - \int v \, du$

$-\frac{x}{x+4} - \int -\frac{dx}{x+4}$

$-\frac{x}{x+4} + \int \frac{dx}{x+4}$

$-\frac{x}{x+4} + \ln|x+4| + C$

5. See attachment

6. An intersting thing happened here

integrating by parts you get ln(x+4) - x/(x+4) + C

By u-substitution I get ln(x+4) +4/(x+4) .

At first I thought I messed up the u sub but the derivative checked out.

but -x/(x+ 4) = 4/(x+4) -1

i.e they differ by a constant so both are right.

7. Thaks so much both of you.