Question

$\int \frac{2x+3}{\sqrt{3-x}}dx$ is equal to (where $C$ is integration constant)

• A. $18\sqrt{3-x}+\frac{2}{3}{(3-x)}^{\frac{3}{2}}+C$
• B. $-18\sqrt{3-x}-\frac{4}{3}{(3-x)}^{\frac{3}{2}}+C$
• C. $-18\sqrt{3-x}+\frac{4}{3}{(3-x)}^{\frac{3}{2}}+C$
• D. $-18\sqrt{3-x}-\frac{2}{3}{(3-x)}^{\frac{3}{2}}+C$

Answer 1

The correct option is C $-18\sqrt{3-x}+\frac{4}{3}{(3-x)}^{\frac{3}{2}}+C$

$I=\int \frac{2x+3}{\sqrt{3-x}}dx$
Put $3-x=t^2$
$\Rightarrow dx=-2tdt$
$\Rightarrow I=\int \frac{2(3-t^2)+3}{t}(-2t)dt=-2\int (9-2t^2)dt=-2[9t-\frac{2}{3}{t}^3]+C=-18\sqrt{3-x}+\frac{4}{3}{(3-x)}^{\frac{3}{2}}+C$