Question

$\int \frac{x+2}{2x^2-7x+3}dx=$

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

Answer 1

Answer: (B)

$\log |\frac{x-3}{\sqrt{2x-1}}|+c$

$\int \frac{x+2}{2x^2-7x+3}dx$
$=\int \frac{x+2}{(x-3)(2x-1)}dx$
$\frac{x+2}{(x-3)(2x-1)}=\frac{A}{(x-3)}+\frac{B}{(2x-1)}$ ....(1)
$\Rightarrow x+2=A(2x-1)+B(x-3)$
Put $x=3\Rightarrow A=1$
Put $x=1\Rightarrow B=-1$
Put this value in (1)
$\frac{x+2}{(x-3)(2x-1)}=\frac{1}{(x-3)}-\frac{1}{(2x-1)}$
$\int \frac{x+2}{(x-3)(2x-1)}dx=\int \frac{1}{(x-3)}dx-\int \frac{1}{(2x-1)}dx$
$=\log |x-3|-\frac{1}{2}\log |2x-1|+C$
$=\log |x-3|-\log \sqrt{2x-1}+C$
$=\log \frac{|x-3|}{\sqrt{2x-1}}+C$