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Q14  Integrate the rational functions \frac{3x-1}{(x+2)^2}

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Given function \frac{3x-1}{(x+2)^2}

can be rewritten as \frac{3x-1}{(x+2)^2} = \frac{A}{(x+2)}+\frac{B}{(x+2)^2}

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

Now, equating the coefficient of x and constant term, we get

A=3 and   2A+B = -1, 

 Solving these equations, we get

B=-7

Therefore, 

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

\implies \int\frac{3x-1}{(x+2)^2}dx = 3 \int \frac{1}{(x+2)}dx-7\int \frac{x}{(x+2)^2}dx

\implies 3\log|x+2| -7\left ( \frac{-1}{(x+2)}\right )+C

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

 

Posted by

Divya Prakash Singh

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