Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# 2 ∫ 1 X + 3 X ( X + 2 ) D X - Mathematics

Sum

$\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx$

#### Solution

$\int_1^2 \frac{x + 3}{x\left( x + 2 \right)} d x$

$= \int_1^2 \frac{x + 2 + 1}{x\left( x + 2 \right)} d x$

$= \int_1^2 \frac{1}{x}dx + \int_1^2 \frac{1}{x\left( x + 2 \right)}dx$

$= \int_1^2 \frac{1}{x}dx + \frac{1}{2} \int_1^2 \frac{\left( x + 2 \right) - x}{x\left( x + 2 \right)}dx$

$= \int_1^2 \frac{1}{x}dx + \frac{1}{2} \int_1^2 \frac{1}{x}dx - \frac{1}{2} \int_1^2 \frac{1}{x + 2}dx$

$= \frac{3}{2} \int_1^2 \frac{1}{x}dx - \frac{1}{2} \int_1^2 \frac{1}{x + 2}dx$

$= \frac{3}{2} \left[ \log x \right]_1^2 - \frac{1}{2} \left[ \log\left( x + 2 \right) \right]_1^2$

$= \frac{3}{2}\log2 - \frac{1}{2}\log4 + \frac{1}{2}\log3$

$= \frac{3}{2}\log2 - \log2 + \frac{1}{2}\log3$

$= \frac{1}{2}\log2 + \frac{1}{2}\log3$

$= \frac{1}{2}\log6$

Concept: Definite Integrals Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 20 Definite Integrals
Revision Exercise | Q 26 | Page 121