Question

$\int \frac{5x+4}{\sqrt{x^2+3x+2}}dx=A\sqrt{x^2+3x+2}-B\ln [(x+\frac{3}{2})+\sqrt{x^2+3x+2}]+C$

Then $A+2B=$?

• A. $9$
• B. $10$
• C. $11$
• D. $12$

Answer 1

Answer: (D)

$12$

Let $I=\int \frac{5x+4}{\sqrt{x^2+3x+2}}dx$
Here, numerator can be written as
$5x+4=A\frac{d}{dx}(x^2+3x+2)+B$
$5x+4=A(2x+3)+B$ .....$\left(1\right)$
$\Rightarrow 5x+4=2Ax+3A+B$
On comparing, we get
$2A=5$ ; $3A+B=4$
Solving these eqns we get
$A=\frac{5}{2};B=-\frac{7}{2}$
Put these values in $\left(1\right)$,
$5x+4=\frac{5}{2}(2x+3)-\frac{7}{2}$
So the given integral becomes
$\int \frac{5x+4}{\sqrt{x^2+3x+2}}dx=\frac{5}{2}\int \frac{2x+3}{\sqrt{x^2+3x+2}}dx-\frac{7}{2}\int \frac{1}{\sqrt{x^2+3x+2}}dx$
Put $x^2+3x+2=t$ in first integral
$\Rightarrow (2x+3)dx=dt$
$=\frac{5}{2}\int \frac{1}{\sqrt{t}}dt-\frac{7}{2}\int \frac{1}{\sqrt{{(x+\frac{3}{2})}^2-{\left(\frac{1}{2}\right)}^2}}dx$
Put $x+\frac{3}{2}=u$
$dx=du$
$I=5\sqrt{t}-\frac{7}{2}\int \frac{1}{\sqrt{u^2-\frac{1}{2^2}}}du$
$=5\sqrt{x^2+3x+2}-\frac{7}{2}\log |u+\sqrt{u^2-{\left(\frac{1}{2}\right)}^2}|+C$
$I=5\sqrt{x^2+3x+2}-\frac{7}{2}\log |x+\frac{3}{2}+\sqrt{(x+\frac{3}{2}{)}^2-{\left(\frac{1}{2}\right)}^2}|+C$
$I=5\sqrt{x^2+3x+2}-\frac{7}{2}\log |x+\frac{3}{2}+\sqrt{x^2+3x+2}|+C$

Hence, $A=5,B=\frac{7}{2}$

$\therefore A+2B=5+7=12$