Question

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

• A. $\frac{11}{4}\log [|x+1||x+3|]+\frac{5}{2(x+1)}+C$
• B. $\frac{11}{4}\log \left|\frac{x+3}{x+1}\right|+\frac{1}{x+1}+C$
• C. $\frac{11}{4}\log |x+2|+\frac{5}{2}(x+3)+\frac{1}{x+1}+C$
• D. $\frac{11}{4}\log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C$

Answer 1

Answer: (D)

Let $l=\int \frac{3x-2}{(x+3){(x+1)}^2}dx$
By partial fraction $\frac{(3x-2)}{(x+3){(x+1)}^2}=\frac{A}{x+3}+\frac{B}{(x+1)}+\frac{C}{{(x+1)}^2}$ ..(i)
$\Rightarrow$ $(3x-2)=A{(x+1)}^2+B(x+3)(x+1)+C(x+3)$
$=A(x^2+1+2x)+B(x^2+4x+3)+C(x+3)$
$=(A+B)x^2+(2A+4B+C)x+(A+3B+3C)$
On comparing the like powers of x, we get
$A+B=0$ .. (ii)
$2A+4B+C=3$ .. (iii)
and $A+3B+3C=-2$ ..(iv)
On multiplying Eq. (iii) by 3and then subtracting it from Eq. (iv), we get $\begin{array}{rl}& 6A+12B+3C=9\\ & A+3B+3C=2\\ & ---+\\ & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ & 5A+9B=11\end{array}$ .. (v)
From Eqs. (ii) and (v), we get
$B=\frac{11}{4}$ and $A=\frac{-11}{4}$
From Eq. (iii), $C=3-2A-4B$
$=3-2\left(\frac{-11}{4}\right)-4\left(\frac{11}{4}\right)$
$=3+\frac{22}{4}-\frac{44}{4}=3-\frac{22}{4}=3-\frac{11}{2}$
$=-\frac{5}{2}$
$\therefore$ From Eq. (i), we get
$\frac{(3x-2)}{(x+3){(x+1)}^2}=\frac{-11}{4(x+3)}+\frac{11}{4(x+1)}.\frac{-5}{2{(x+1)}^2}$
$\Rightarrow$ $\int \frac{(3x-2)}{(x+3){(x+1)}^2}dx$
$=\int \left\{\frac{-11}{4(x+3)}+\frac{11}{4(x+1)}-\frac{5}{2{(x+1)}^2}\right\}dx$
$=-\frac{11}{4}\log |x+3|+\frac{11}{4}\log |x+1|+\frac{5}{2(x+1)}+C$
$=\frac{11}{4}\log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C$