Question

Evaluate: ${\displaystyle \int \frac{2\text{x}+1}{\left(\text{x + 1}\right)\left(\text{x - 2}\right)}}$ dx

Answer 1

Let I = ${\displaystyle \int \frac{2\text{x}+1}{\left(\text{x + 1}\right)\left(\text{x - 2}\right)}}$ dx

Let ${\displaystyle \frac{2\text{x}+1}{\left(\text{x + 1}\right)\left(\text{x - 2}\right)}=\frac{\text{A}}{\text{x + 1}}+\frac{\text{B}}{\text{x - 2}}}$

∴ 2x + 1 = A(x - 2) + B(x + 1)   ....(i)

Putting x = - 1 in (i), we get

2(-1) + 1 = A(- 3) + B(0)

∴ - 1 = -3A

∴ A = ${\displaystyle \frac{1}{3}}$

Putting x = 2 in (i), we get

2(2) + 1 = A(0) + B(3)

∴ 5 = 3B

∴ B = ${\displaystyle \frac{5}{3}}$

∴ ${\displaystyle \frac{2\text{x}+1}{\left(\text{x + 1}\right)\left(\text{x - 2}\right)}=\frac{\frac{1}{3}}{\text{x + 1}}+\frac{\frac{5}{3}}{\text{x - 2}}}$

∴ I = ${\displaystyle \int (\frac{\left(\frac{1}{3}\right)}{\text{x + 1}}+\frac{\left(\frac{5}{3}\right)}{\text{x - 2}})}$ dx

∴ ${\displaystyle \frac{1}{3}\int \frac{1}{\text{x + 1}}\text{dx}+\frac{5}{3}\int \frac{1}{\text{x - 2}}}$ dx

∴ I = ${\displaystyle \frac{1}{3}\log |\text{x}+1|+\frac{5}{3}\log \left|\text{x - 2}\right|+\text{c}}$