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Question

Mathematics Question on integral

Integrate the rational function: 3x−1(x+2)2\frac {3x-1}{(x+2)^2}

Answer

Let 3x−1(x+2)2\frac {3x-1}{(x+2)^2} == A(x+2)+B(x+2)2\frac {A}{(x+2)}+\frac {B}{(x+2)^2}

⇒3x−1=A(x+2)+B⇒ 3x-1 = A(x+2)+B

Equating the coefficient of x and constant term, we obtainEquating\ the \ coefficient \ of\ x \ and \ constant\ term, \ we \ obtain
A=3A = 3
2A+B=−1⇒B=−72A + B = −1 ⇒ B = −7

∴ 3x−1(x+2)2\frac {3x-1}{(x+2)^2} == 3(x+2)−7(x+2)2\frac {3}{(x+2)} - \frac {7}{(x+2)^2}

⇒ ∫$$\frac {3x-1}{(x+2)^2}\ dx == 3∫1(x+2)dx−7∫x(x+2)2dx3∫\frac {1}{(x+2)}dx - 7∫\frac {x}{(x+2)^2}dx

                          $= 3log\ |x+2|-7(\frac {-1}{(x+2)})+C$

                          $= 3log\ |x+2|+\frac {7}{(x+2)} +C$