integral from 1 to infinity of $\frac{5}{(4x+2)^3}$ I have solved the integral:
$$\int_1^\infty{\frac{5}{(4x+2)^3}}dx$$
using u substitution and I am not getting the correct answer. I am missing some step here or making an algebra error. I am not sure what is wrong. The work yields a finite answer (I am doing the problem for a integral test on a series), just the wrong one. Could someone work through the steps?
I first moved 5 out as a constant. I then supposed u=4x+2 and got 1/4du=dx. That left me with the integral of 1/4(u^3)du from 6 to infinity (changing the bounds to u by plugging the original bounds into u=4x+2) I moved 1/4 out as a constant, leaving me with 5/4 as a constant. I integrated to 5/4(-1/4(u^4) evaluated from 6 to infinity). I solve that and get -5/20736.
 A: This is a simple case of confusion over how to integrate a positive power in the denominator.  The antiderivative of $1/u^3$ is $-1/(2u^2)$, not $-1/(4u^4)$.  I.e.,
$$\int {du\over u^3}={-1\over2u^2}+C$$
or
$$\left(1\over u^2\right)'=(u^{-2})'=-2u^{-3}$$
A: To solve the problem $$\int_1^\infty{\frac{5}{(4x+2)^3}}\mathrm dx,$$
first substitute for $u = 4x+2$ and $\mathrm du = 4 \space \mathrm dx$. This gives $$\frac{5}{4}\int_1^\infty{\frac{1}{u^3}\mathrm du}.$$
Since $$\frac{\mathrm d}{\mathrm du}\left[-\frac{1}{2u^2}\right] = \frac{1}{u^3},$$
the integral will evaluate to $$\frac{5}{4}\left[-\frac{1}{2u^2}\right]^{\infty}_{x=1},$$
and substitute for $x$ to get
$$\frac{5}{4}\left[-\frac{1}{2(4x+2)^2}\right]^{\infty}_1 = \frac{5}{4}\left(\frac{1}{72}\right) = \frac{5}{288}.$$
QED
A: we suppose that u=4x+2
du=4dx , therefore dx=1/4 (du)
A: I first moved 5 out as a constant. I then supposed u=4x+2 and got 1/4du=dx. That left me with the integral of 1/4(u^3)du from 6 to infinity (changing the bounds to u by plugging the origional bounds into u=4x+2) I moved 1/4 out as a constant, leaving me with 5/4 as a constant. I integrated to 5/4(-1/4(u^4) evaluated from 6 to infinity). I solve that and get -5/20736.  
