Question

$\int \frac{8x+13}{\sqrt{4x+7}}dx$ is equal to
(where $C$ is constant of integration)

• A. $\frac{2}{3}(4x+7{)}^{\frac{3}{2}}-\frac{1}{2}(4x+7{)}^{\frac{1}{2}}+C$
• B. $\frac{1}{3}(4x+7{)}^{\frac{3}{2}}-\frac{1}{2}(4x+7{)}^{\frac{1}{2}}+C$
• C. $\frac{1}{2}(4x+7{)}^{\frac{3}{2}}-\frac{1}{3}(4x+7{)}^{\frac{1}{2}}+C$
• D. $\frac{1}{3}(4x+7{)}^{\frac{3}{2}}-\frac{1}{3}(4x+7{)}^{\frac{1}{3}}+C$

Answer 1

The correct option is B $\frac{1}{3}(4x+7{)}^{\frac{3}{2}}-\frac{1}{2}(4x+7{)}^{\frac{1}{2}}+C$

$\int \frac{8x+13}{\sqrt{4x+7}}dx$
$=\int \frac{8x+14-1}{\sqrt{4x+7}}dx$
$=\int \frac{2(4x+7)-1}{\sqrt{4x+7}}dx$
$=2\int \sqrt{4x+7}dx-\int \frac{1}{\sqrt{4x+7}}dx$
$=2\left\{\frac{(4x+7{)}^{\frac{3}{2}}}{4\times \frac{3}{2}}\right\}-\left\{\frac{(4x+7{)}^{\frac{1}{2}}}{4\times \frac{1}{2}}\right\}+C$
$=\frac{1}{3}(4x+7{)}^{\frac{3}{2}}-\frac{1}{2}(4x+7{)}^{\frac{1}{2}}+C$