Question

Find : \[ \int \frac{2x+1}{\sqrt{x^2+6x}}\,dx \]

Hint:

When the numerator is not exactly the derivative of the expression inside the square root, rewrite the numerator in terms of that derivative. This allows the integral to be split and solved using substitution and standard formulas.

Answer 1

Step 1: Observe the expression inside the square root.
The denominator contains the expression \[ \sqrt{x^2+6x} \] First differentiate the expression inside the square root: \[ \frac{d}{dx}(x^2+6x)=2x+6 \] The numerator of the integral is \(2x+1\), which is similar to the derivative \(2x+6\). Therefore, we adjust the numerator algebraically.
Step 2: Rewrite the numerator.
\[ 2x+1 = (2x+6)-5 \] Thus the integral becomes \[ \int \frac{2x+6}{\sqrt{x^2+6x}}dx - 5\int \frac{1}{\sqrt{x^2+6x}}dx \] Step 3: Solve the first integral using substitution.
Let \[ t = x^2+6x \] Then \[ dt = (2x+6)dx \] Thus \[ \int \frac{2x+6}{\sqrt{x^2+6x}}dx = \int \frac{dt}{\sqrt{t}} \] \[ =2\sqrt{t} \] Substitute back: \[ 2\sqrt{x^2+6x} \] Step 4: Solve the second integral.
\[ \int \frac{1}{\sqrt{x^2+6x}}dx \] Complete the square: \[ x^2+6x = (x+3)^2-9 \] Thus \[ \int \frac{1}{\sqrt{(x+3)^2-9}}dx \] Using the standard formula \[ \int \frac{dx}{\sqrt{x^2-a^2}}=\ln\left|x+\sqrt{x^2-a^2}\right| \] Therefore \[ \int \frac{dx}{\sqrt{(x+3)^2-9}} = \ln\left|x+3+\sqrt{x^2+6x}\right| \] Step 5: Combine the results.
\[ \int \frac{2x+1}{\sqrt{x^2+6x}}dx = 2\sqrt{x^2+6x}-5\ln\left|x+3+\sqrt{x^2+6x}\right|+C \] Final Answer:
\[ \boxed{2\sqrt{x^2+6x}-5\ln\left|x+3+\sqrt{x^2+6x}\right|+C} \]