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) \[\frac{11}{4}\log \,\left| \frac{x+1}{x+3} \right|+\frac{5}{2(x+1)}+C\]
Correct Answer: D
Solution :
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 \[\frac{(3x-2)}{(x+3)\,{{(x+1)}^{2}}}=\frac{-11}{4(x+3)}+\frac{11}{4(x+1)}.\frac{-5}{2{{(x+1)}^{2}}}\] \[\Rightarrow \] \[\int{\frac{(3x-2)}{(x+3)\,{{(x+1)}^{2}}}dx}\] \[=\int{\left\{ \frac{-11}{4(x+3)}+\frac{11}{4(x+1)}-\frac{5}{2{{(x+1)}^{2}}} \right\}dx}\] \[=-\frac{11}{4}\log \,|x+3|+\frac{11}{4}\,\log |x+1|+\frac{5}{2(x+1)}+C\] \[=\frac{11}{4}\log \left| \frac{x+1}{x+3} \right|+\frac{5}{29x+1)}+C\]
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