Question

Find the integrals (primitives):
$\rm \displaystyle\int \frac{3x-7}{x+1} \ dx$

Answer 1

Such type of expression can be easily integrated with the help of substitution Method of integration

$let \: x + 1 = t \\ \\ dx = dt \\ \\ x = t - 1$
$\int \frac{3x-7}{x+1} \ dx = \int \: \frac{3(t - 1) - 7}{t} dt \\ \\ = \int \: \frac{3t - 3 - 7}{t} dt \\ \\ = \int \frac{3t}{t} \: dt - \int \: \frac{10}{t} dt \\ \\ = \int \: 3 \: dt \: - 10\int \: \frac{1}{t} dt \\ \\ = 3t - 10 \: log \: t + C$
redo substitution

$\int \frac{3x-7}{x+1} \ dx= 3(x + 1)- 10 \: log \: (x + 1)+ C$
Hope it helps you.