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 $\frac{3-\sqrt{5}x}{1+\sqrt{5}x}=u$,  we have $\frac{-4\sqrt{5}} 
{(1+\sqrt{5}x)^2} dx=du$ thus
$$\int_{0}^{1} {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx=\frac{1}{-4\sqrt{5}}\int_{3}^{\sqrt{5}-2}u\,du=\frac 12$$
