Prove $\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx={1\over 2}$ using an alternative method Prove that
$$\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx={1\over 2}\tag1$$
My try:
$u=1+\sqrt{5}x$ then $du=\sqrt{5} \, dx$
$${1\over \sqrt 5}\int_1^{1+\sqrt{5}}(4u^{-3}-u^{-2}) \, du$$
$$\left. {1\over \sqrt{5}}(-2u^{-2}+u^{-1}) \right|_1^{1+\sqrt{5}}={1\over 2}$$
Prove $(1)$ using an alternative method other than substitution method.

Set $x=\frac{\sinh^2u}{\sqrt{5}}$ and $dx=\frac{2\sinh u\cosh u}{\sqrt{5}}$
So $$\int{3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx=\int \frac{6\sinh u\cosh u}{\sqrt{5}\cosh^6 u}du -\int\frac{2\sinh^3u \cosh u}{\sqrt{5}\cosh^6 u} \, du = \int \frac{6\sinh u}{\sqrt{5}\cosh^5 u} \, du -\int \frac{2(1+\cosh^2 u ) \sinh u}{\sqrt{5}\cosh^5 u} \, du$$
And you can calculate these integrals using the substitution $t=\cosh u$ and $dt=\sinh u \, du$
Another way using partial fractions: 
$$\int{3-\sqrt{5}x\over (1+\sqrt{5}x)^3} dx= \int \frac{20\sqrt{5}}{(5x+\sqrt{5})^3}- \frac{5}{(5x+\sqrt{5})^2}dx $$
