Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals3 Marks
Question
Evaluate the following intregals: $\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}$
✓
Answer
Let $\text{I}=\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}$
We express
$\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{\text{x}+3}$
$\Rightarrow3\text{x}-2=\text{A}(\text{x}+1)(\text{x}+3)+\text{B}(\text{x}+3)+\text{C}(\text{x}+1)^2$
Equating the coefficient of $x^2, x$ and constants, we get
$0 = A + C$ and $3 = 4A + B + 2C$ and $-2 = 3A + 3B + C$
$\text{or }\text{A}=\frac{11}{4}\text{ and }\text{B}=-\frac{5}{2}\text{ and }\text{C}=-\frac{11}{4}$
$\therefore\text{I}=\int\bigg(\frac{\frac{11}{4}}{\text{x}+1}+\frac{-\frac{5}{2}}{(\text{x}+1)^2}+\frac{-\frac{11}{4}}{\text{x}+3}\bigg)\ \text{dx}$
$=\frac{11}{4}\int\frac{1}{\text{x}+1}\text{ dx }-\frac{5}{2}\int\frac{1}{(\text{x}+1)^2}\text{ dx }-\frac{11}{4}\int\frac{1}{\text{x}+3}\text{ dx}$
$=\frac{11}{4}\log|\text{x}+1|+\frac{5}{2(\text{x}+1)}-\frac{11}{4}\log|\text{x}+3|+\text{C}$
Hence,
$\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}$
$=\frac{11}{4}\log|\text{x}+1|+\frac{5}{2(\text{x}+1)}-\frac{11}{4}\log|\text{x}+3|+\text{C}$
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