Integrate x / (x+1)^3

Oct 2015
27
0
Canada
Hello,

Could you help me with this?

Integrate x / (x+1)^3 from a=1 to b=4.

I don't think that substitution works here, so my idea was to approach this using integration by parts.
I tried solving it with u=(x+1)^-3 and dv = x, but that didn't work out because I ended up with something that I couldn't integrate.
Should u=x and dv=(x+1)^-3?

Then du=1 and v = ???

What's the correct method to use to solve this and what steps do I follow?
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Substitution DOES work here and simplifies the problem greatly. Let $\displaystyle \begin{align*} u = x + 1 \implies \mathrm{d}u = \mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{x}{ \left( x + 1 \right) ^3}\,\mathrm{d}x} &= \int{ \frac{ u - 1}{u^3} \,\mathrm{d}u } \\ &= \int{ u^{-2} - u^{-3}\,\mathrm{d}u } \\ &= \frac{ u^{-1}}{-1} - \left( \frac{u^{-2}}{-2} \right) + C \\ &= -\frac{1}{u} + \frac{1}{2u^2} + C \\ &= \frac{1 - 2u}{2u^2} + C \\ &= \frac{1 - 2 \left( x + 1 \right) }{ 2\left( x + 1 \right) ^2} + C \end{align*}$
 
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Oct 2015
27
0
Canada
Thank you for your quick reply. I tried substitution with u=x+1, but before now, I didn't know that I could replace the x in the numerator by seeing that x=u-1.