You can also do it like this: $$\frac{1}{x(x+1)}=\frac1x-\frac{1}{x+1}$$ Multiply by $1/x$: $$\frac{1}{x^2(x+1)}=\frac1{x^2}-\frac{1}{x(x+1)}=\frac1{x^2}-\frac1x+\frac1{x+1}$$ and again $$\frac{1}{x^2(x+1)}=\frac1{x^3}-\frac1{x^2}+\frac1{x(x+1)}=\frac{1}{x^3}-\frac{1}{x^2}+\frac1x-\frac{1}{x+1}$$ See the pattern and you can prove $$\boxed{\frac{1}{x^n(x+1)}=\frac{1}{x^n}-\frac{1}{x^{n-1}}+\cdots-(-1)^n\frac1x+(-1)^n\frac{1}{x+1}}$$ Set $n=4$ and get $$\frac1{x^4(x+1)}=\frac1{x^4}-\frac1{x^3}+\frac1{x^2}-\frac1x+\frac1{x+1}$$ Hope this helps. :)

You can also do it like this: $$\frac{1}{x(x+1)}=\frac1x-\frac{1}{x+1}$$ Multiply by $1/x$: $$\frac{1}{x^2(x+1)}=\frac1{x^2}-\frac{1}{x(x+1)}=\frac1{x^2}-\frac1x+\frac1{x+1}$$ and again $$\frac{1}{x^3(x+1)}=\frac1{x^3}-\frac1{x^2}+\frac1{x(x+1)}=\frac{1}{x^3}-\frac{1}{x^2}+\frac1x-\frac{1}{x+1}$$ See the pattern and you can prove $$\boxed{\frac{1}{x^n(x+1)}=\frac{1}{x^n}-\frac{1}{x^{n-1}}+\cdots-(-1)^n\frac1x+(-1)^n\frac{1}{x+1}}$$ Set $n=4$ and get $$\frac1{x^4(x+1)}=\frac1{x^4}-\frac1{x^3}+\frac1{x^2}-\frac1x+\frac1{x+1}$$ Hope this helps. :)