Question

$\int \frac{2x+3}{\sqrt{4x+3}}dx=$

• A. $\frac{1}{12}(4x-3{)}^{\frac{3}{2}}+\frac{1}{4}\sqrt{4x+3}+c$
• B. $\frac{1}{12}(4x+3{)}^{\frac{3}{2}}+\frac{3}{4}\sqrt{4x+3}+c$
• C. $\frac{1}{12}(4x+3{)}^{\frac{3}{2}}-\frac{3}{4}\sqrt{4x-3}+c$
• D. $\frac{1}{12}(4x+3{)}^{\frac{3}{2}}-\frac{1}{4}\sqrt{4x-3}+c$

Answer 1

The correct option is B $\frac{1}{12}(4x+3{)}^{\frac{3}{2}}+\frac{3}{4}\sqrt{4x+3}+c$

$\int \frac{2x+3}{\sqrt{4x+3}}dx$
$=\frac{1}{2}\int \frac{4x+3+3}{\sqrt{4x+3}}dx$
$=\frac{1}{2}\int \sqrt{4x+3}dx+\frac{1}{2}\int \frac{3}{\sqrt{4x+3}}dx$
Put $4x+3=t$
$\Rightarrow dx=\frac{dt}{4}$
$=\frac{1}{8}\int \sqrt{t}dt+\frac{3}{8}\int \frac{1}{\sqrt{t}}dt$
$=\frac{1}{12}{t}^{3/2}+\frac{3}{4}{t}^{1/2}+C$
$=\frac{1}{12}(4x+3{)}^{3/2}+\frac{3}{4}\sqrt{4x+3}+C$