Question

$\int \frac{8x+13}{\sqrt{4x+7}}dx$ is equal to :

• A. $\frac{1}{6}(8x+11)\sqrt{4x+7}+c$
• B. $\frac{1}{6}(8x+13)\sqrt{4x+7}+c$
• C. $\frac{1}{6}(8x+9)\sqrt{4x+7}+c$
• D. $\frac{1}{6}(8x+15)\sqrt{4x+7}+c$

Answer 1

Answer: (A)

$\frac{1}{6}(8x+11)\sqrt{4x+7}+c$

Let $I=\int \frac{8x+13}{\sqrt{4x+7}}dx$
Put $t=\sqrt{4x+7}\Rightarrow dt=\frac{2}{\sqrt{4x+7}}dx$
$I=\frac{1}{2}\int (2(t^2-7)+13)dt=\int t^{2}dt-\int 7dt+\int \frac{13}{2}dt$
$=\int t^{2}dt-\frac{1}{2}\int dt=\frac{t^3}{3}-\frac{t}{2}+c$
$=\frac{(\sqrt{4x+7}{)}^3}{3}-\frac{\sqrt{4x+7}}{2}+c=\frac{1}{6}(8x+11)\sqrt{4x+7}+c$