1. Integrate

How to integrate this? I = (10x-2x^2)/[(x-1)^2 * (x+3)] dx

2. Re: Integrate

Hint: convert the expression using partial fraction expansion into three separate fractions:

$\frac {10x-2x^2}{(x-1)^2(x+3)} = \frac A {(x-1)^2} + \frac B {x-1} + \frac C {x+3}$

Determine values for A, B and C and then it's easy to integrate.

3. Re: Integrate

By partial fraction expansion we know that:

\begin{align*} \frac A {(x-1)^2} + \frac B {x-1} + \frac C {x+3} = \frac {10x-2x^2}{(x-1)^2(x+3)} \\A(x+3) + B(x+3)(x-1) + C(x-1)^2 =& -2x^2 +10x\\ (B+C)x^{2} +(A +2B - 2C)x + (3A - 3B + C) =& -2x^2 +10x\end{align*}

\begin{align*}B + C =& -2.....\text{(1)}\\ A + 2B - 2C =& 10........\text{(2)}\\ 3A - 3B + C =& 0..........\text{(3)}\end{align*}

Solving these three equations we get:

$\therefore\,\, A = 2, \,\, B = 1 \,\, \text{and } C = -3$

Now:

\begin{align*}\int \frac {10x-2x^2}{(x-1)^2(x+3)}\,\,dx =& \int \left(\frac{2}{(x-1)^2} + \frac{1}{x-1} - \frac{3}{x+3}\right)\,\,dx\\=& \int \frac{2}{(x-1)^2}\,\,dx + \int \frac{1}{x-1}\,\,dx - \int \frac{3}{x+3}\,\,dx....\text{(4)}\end{align*}

$\text{Let } u = x - 1 \therefore\,\, du = dx$

$\text{ and let } v = x + 3 \therefore\,\, dv = dx$

\begin{align*}\int \frac {10x-2x^2}{(x-1)^{2}(x+3)}\,\,dx =& \int \frac{2}{u^2}\,\,du +\int \frac{1}{u}\,\,du -\int\frac{3}{v}\,\,dv.............\text{[by plugging in eq.4 }u, v, du, dv\text{]}\\ =& \frac{-2}{u} + \ln{(u)} - 3\ln{(v)} + C\\ =& \frac{-2}{x-1} + \ln{(x-1)} - 3\ln{(x+3)} + C...\text{[by plugging in values of } u \text{ and } v \text{]}\end{align*}