Question

Find the integral of the function $\frac{1}{\sqrt{4x^2+64x+100}}$

• A. $ln|(x+8)+\sqrt{(x+8{)}^2+{10}^2}|+c$
• B. $ln|(x+4)+\sqrt{(x+4{)}^2+3^2}|+c$
• C. $ln|(x+8)+\sqrt{(x+8{)}^2+{10}^2}|+c$
• D. $ln|(x+8)+\sqrt{(x+4{)}^2+3^2}|+c$

Answer 1

The correct option is B $ln|(x+4)+\sqrt{(x+4{)}^2+3^2}|+c$

$\int \frac{dx}{\sqrt{4x^2+64x+100}}$
We will try to express this as a perfect square and proceed
$\int \frac{dx}{\sqrt{4x^2+64x+100}}=\int \frac{dx}{\sqrt{4(x^2+16x+25)}}$
=$\int \frac{dx}{2\sqrt{(x+4{)}^2+9)}}$
This integral is of the form $\int \frac{dx}{\sqrt{x^2+a^2}}$ which is equal to $ln|x+\sqrt{x^2+a^2}|+c$
$\Rightarrow \int \frac{dx}{2\sqrt{(x+4{)}^2+3^2)}}=ln|(x+4)+\sqrt{(x+4{)}^2+3^2}|+c$