How to solve $\int \frac{x^3}{(1+x)^5}\ dx$?

Question

I am stuck at this integral. Can anybody help or give me a hint what to do with the denominator?

$$\int \frac{x^3}{(1+x)^5} \, dx$$

Answer 1

Try to integrate by parts:

$$ \int x^2\cdot (1+x)^{-5}\ dx=x^2\cdot \left(-\frac{1}{4}(1+x)^{-4}\right)+\int 2x\cdot \frac{1}{4}(1+x)^{-4}\ dx $$

and then apply again the rule:

$$ \int 2x\cdot \frac{1}{4}(1+x)^{-4}\ dx= \frac{1}{2}x\left(-\frac{1}{3}(1+x)^{-3}\right)+\int \frac{1}{2}\cdot \frac{1}{3}(1+x)^{-3}\ dx $$

and the last integral is immediate

$$ \int \frac{1}{2}\cdot \frac{1}{3}(1+x)^{-3}\ dx=-\frac{1}{12}(1+x)^{-2}+cost $$

so, summing up all the terms, you get

$$ \int \frac{x^2}{(1+x)^5}\ dx=-\frac{6x^2+4x+1}{12(1+x)^4}+cost. $$

Answer 2

Note that\begin{align}\int\frac{x^3}{(1+x)^5}\,\mathrm dx&=\int\frac{((1+x)-1)^3}{(1+x)^5}\,\mathrm dx\\&=\int(1+x)^{-2}\,\mathrm dx-3\int(1+x)^{-3}\,\mathrm dx+\\&\phantom{=}+3\int(1+x)^{-4}\,\mathrm dx-\int(1+x)^{-5}\,\mathrm dx.\end{align}Besides$$n\ne-1\implies\int(1+x)^n\,\mathrm dx=\frac1{n+1}(1+x)^{n+1}.$$

Answer 3

Let $$1+x = 1/t$$ $$dx = \frac {-1} {t^2} dt$$ The integral transforms into $$\int \frac {(\frac 1 t -1)^3 } {(\frac 1 t)^5} \frac {-dt} {t^2}$$ $$\int (t-1)^3 dt$$ which can be easily solved further

Tip : Whenever the integral contains linear terms with higher powers in the denominator, substituting the linear term with 1/t becomes useful for solving the integral