Question

Find: \[ \int \frac{3x+5}{\sqrt{x^2+2x+4}} \,dx. \]

Hint:

For integrals with quadratic expressions inside square roots, complete the square to simplify.

Answer 1

Step 1: Denominator Completion of the Square
The quadratic expression in the denominator is: \[ x^2 + 2x + 4. \] Completing the square yields: \[ x^2 + 2x + 4 = (x+1)^2 + 3. \]
Step 2: Substitution Application
Let: \[ t = x+1 \quad \text{therefore} \quad dt = dx. \] The integral transforms to: \[ I = \int \frac{3(t-1) + 5}{\sqrt{t^2 + 3}} \, dt. \] Simplifying the numerator: \[ 3(t-1) + 5 = 3t - 3 + 5 = 3t + 2. \] The integral becomes: \[ I = \int \frac{3t + 2}{\sqrt{t^2 + 3}} \, dt. \]
Step 3: Integral Decomposition
The expression is split as: \[ I = \int \frac{3t}{\sqrt{t^2+3}} \,dt + \int \frac{2}{\sqrt{t^2+3}} \,dt. \]
First Integral: Using the substitution \( u = t^2 + 3 \), which implies \( du = 2t \, dt \): \[ \int \frac{3t}{\sqrt{t^2+3}} \, dt = \frac{3}{2} \int \frac{du}{\sqrt{u}}. \] The integral \( \int u^{-1/2} \, du \) evaluates to \( 2u^{1/2} \), resulting in: \[ \frac{3}{2} \cdot 2\sqrt{t^2+3} = 3\sqrt{t^2+3}. \]
Second Integral: Applying the standard formula \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \sinh^{-1} \left( \frac{x}{a} \right) \): \[ \int \frac{2}{\sqrt{t^2+3}} \,dt = 2\sinh^{-1} \left(\frac{t}{\sqrt{3}}\right). \]
Step 4: Back-Substitution \( t = x+1 \)
Substituting \( t \) back in terms of \( x \): \[ I = 3\sqrt{x^2+2x+4} + 2\sinh^{-1} \left(\frac{x+1}{\sqrt{3}}\right) + C. \]
Final Answer: \[ \int \frac{3x+5}{\sqrt{x^2+2x+4}} \,dx = 3\sqrt{x^2+2x+4} + 2\sinh^{-1} \left(\frac{x+1}{\sqrt{3}}\right) + C. \]