Question

Find the integrals (primitives):
$\rm \displaystyle\int \frac{2x^{2}-4x-5}{x-1} \ dx$

Answer 1

Integration by substitution Method:

$let \: x - 1 = t \\ \\ dx = dt \\ \\ x = t + 1$

$\int \frac{2x^{2}-4x-5}{x-1} \ dx \\ \\ \int \frac{2(t + 1)^{2}-4(t + 1) - 5}{t} \ dt \\ \\ \int \frac{2 {t}^{2} + 2 + 4t -4t - 4 - 5}{t} \ dt \\ \\ \int \frac{2 {t}^{2} - 7}{t} \ dt \\ \\ \int \: 2 \: t \: dt - 7\int \: \frac{1}{t} dt \\ \\ = 2 \frac{ {t}^{2} }{2} - 7 \: log \: t + C \\ \\ = {t}^{2} - 7 \: log \: t + C$
redo substitution

$\int \frac{2x^{2}-4x-5}{x-1} \ dx= {(x - 1)}^{2} - 7 \: log \: |x - 1|+ C$
Hope it helps you.