Question

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

• A. $\frac{2}{5}\log |2x+1|-\frac{11}{15}\log |1-3x|+c$
• B. $\frac{2}{5}\log |2x+1|-\frac{11}{15}\log |1+3x|+c$
• C. $\frac{2}{3}\log |2x-1|+\frac{11}{15}\log |1-3x|+c$
• D. $\frac{2}{5}\log |2x+1|-\frac{21}{5}\log |1+3x|+c$

Answer 1

The correct option is A $\frac{2}{5}\log |2x+1|-\frac{11}{15}\log |1-3x|+c$

We can express the numerator as,
$2x+3=A(2x+1)+B(1-3x)$
Comparing the corresponding coefficients,
A = $\frac{11}{5}$ and $B=\frac{4}{5}$
Thus,
$\int \frac{2x+3}{(2x+1)(1-3x)}dx=\int \frac{A(2x+1)dx+B(1-3x)}{(2x+1)(1-3x)}dx$
$=\int \frac{A}{1-3x}dx+\int \frac{B}{2x+1}dx$
$=\frac{-11}{15}\times \ln (1-3x)+\frac{2}{5}\times \ln (2x+1)+c$
Hence, option 'A' is correct.