∫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.
∫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.
$\displaystyle u = x$ ... $\displaystyle dv = \frac{1}{(x+4)^2} \, dx$
$\displaystyle du = dx$ ... $\displaystyle v = -\frac{1}{x+4}$
$\displaystyle uv - \int v \, du$
$\displaystyle -\frac{x}{x+4} - \int -\frac{dx}{x+4}$
$\displaystyle -\frac{x}{x+4} + \int \frac{dx}{x+4}$
$\displaystyle -\frac{x}{x+4} + \ln|x+4| + C$
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.