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{align} & 6A+12B+3C=9 \\ & A+3B+3C=2 \\ & -\,\,\,\,\,-\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,+ \\ & \_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ & 5A+9B=11 \\ \end{align} $ .. (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 $