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.

Using integration by parts : $$\begin{align}\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx &= \int_0^1 \left( - \frac{1}{2\sqrt{5} (1+\sqrt{5}x)^2} \right)'(3-\sqrt{5}x) \, dx \\[8pt] &= \left[ \frac{\sqrt{5}x - 3}{2\sqrt{5}(1+\sqrt{5}x)^2}\right]_0^1 - \frac{1}{2} \int_0^1 \frac{dx}{(1+\sqrt{5}x)^2} \\[8pt] &=  \left[ \frac{\sqrt{5}x - 3}{2\sqrt{5}(1+\sqrt{5}x)^2} + \frac{1}{10x + 2\sqrt{5}}\right]_0^1 \\[8pt] &= \frac{1}{2} \end{align}  $$
