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: Complete the square for the denominator
The expression inside the square root is \(x^2 + 6x\). Completing the square: \[ x^2 + 6x = (x+3)^2 - 9. \] So, the integral becomes: \[ \int \frac{2x+1}{\sqrt{(x+3)^2 - 9}}\,dx. \]

Step 2: Substitution
Let \( u = x + 3 \), so that \( du = dx \). Substituting into the integral: \[ \int \frac{2(x+3) - 5}{\sqrt{(x+3)^2 - 9}}\,dx = \int \frac{2u - 5}{\sqrt{u^2 - 9}}\,du. \] Now, separate the integral: \[ \int \frac{2u}{\sqrt{u^2 - 9}}\,du - 5\int \frac{1}{\sqrt{u^2 - 9}}\,du. \]

Step 3: Solve the first integral
For the first integral, use the substitution \( v = u^2 - 9 \) with \( dv = 2u\,du \), which simplifies to: \[ \int \frac{2u}{\sqrt{u^2 - 9}}\,du = \sqrt{u^2 - 9}. \]

Step 4: Solve the second integral
For the second integral, use the standard result: \[ \int \frac{1}{\sqrt{u^2 - 9}}\,du = \text{arcosh}\left(\frac{u}{3}\right). \]

Step 5: Final solution
Substituting back \( u = x + 3 \) into both integrals: \[ \sqrt{(x+3)^2 - 9} - 5 \, \text{arcosh}\left(\frac{x+3}{3}\right) + C. \]

Final Answer:
\[ \sqrt{x^2 + 6x + 9 - 9} - 5 \, \text{arcosh}\left(\frac{x+3}{3}\right) + C \] \[ = \sqrt{x^2 + 6x} - 5 \, \text{arcosh}\left(\frac{x+3}{3}\right) + C. \]