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.
 A: Using integration by parts : $$\begin{align}\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx &= \int_0^1 \left( - \frac{1}{2\sqrt{5} (1+\sqrt{5}x)^2} \right)'(3-\sqrt{5}x) \, dx \\[8pt] &= \left[ \frac{\sqrt{5}x - 3}{2\sqrt{5}(1+\sqrt{5}x)^2}\right]_0^1 - \frac{1}{2} \int_0^1 \frac{dx}{(1+\sqrt{5}x)^2} \\[8pt] &=  \left[ \frac{\sqrt{5}x - 3}{2\sqrt{5}(1+\sqrt{5}x)^2} + \frac{1}{10x + 2\sqrt{5}}\right]_0^1 \\[8pt] &= \frac{1}{2} \end{align}  $$
A: Set $\frac{3-\sqrt{5}x}{1+\sqrt{5}x}=u$,  we have $\frac{-4\sqrt{5}} 
{(1+\sqrt{5}x)^2} dx=du$ thus
$$\int_{0}^{1} {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx=\frac{1}{-4\sqrt{5}}\int_{3}^{\sqrt{5}-2}u\,du=\frac 12$$
A: 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: 
$$\int{3-\sqrt{5}x\over (1+\sqrt{5}x)^3} dx= \int \frac{20\sqrt{5}}{(5x+\sqrt{5})^3}- \frac{5}{(5x+\sqrt{5})^2}dx $$
A: $$
{3-\sqrt{5}x\over (1+\sqrt 5 \, x)^3} = \frac A {1+\sqrt 5\,x} + \frac B {(1+\sqrt 5\,x)^2} + \frac C {(1+\sqrt 5\,x)^3}
$$
Find $A$, $B$, and $C$ and then integrate each term separately.
