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# Evaluate : $\int\large\frac{(3x-2)}{(x+1)^2(x+3)} $$dx Can you answer this question? ## 1 Answer 0 votes Toolbox: • Form of the rational function \large\frac{px+q}{(x^2+a)(x^2+b)} • Form of the partial function \large\frac{Ax+B}{(x^2+a)}+\frac{Cx+D}{(x^2+b)} • \int\large\frac{dx}{x}$$=log|x|+c.$
Step 1:
$I=\int\large\frac{(3x-2)}{(x+1)^2(x+3)^2}$$dx Consider \large\frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3} \Rightarrow (3x-2)=A(x+1)(x+3)+B(x+3)+C(x+1)^2 \Rightarrow A(x^2+4x+3)+B(x+3)+C(x^2+2x+1) Step 2: Now compare the coefficient of like terms, For x^2 \quad A+C=0 For x \quad 4A+B+2C=3 For the constant term \quad 3A+3B+C=-2 Solving the above equations we get, A=\large\frac{11}{4}$$B=\large\frac{-5}{2}$ and $C=\large\frac{-11}{4}$
Step 3:
$\large\frac{3x-2}{(x+1)^2(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^2}-\frac{11}{4(x+3)}$
$5=\large\frac{11}{4}\int\frac{dx}{x+1}-\frac{5}{2}\int\frac{dx}{(x+1)^2}-\frac{11}{4}\int\frac{dx}{x+3}$
On integrating we get,
$I=\large\frac{11}{4}$$\log\mid x+1\mid+\large\frac{5}{2(x+1)}-\frac{11}{4}$$\log\mid x+3\mid+c$