Question

Evaluate the integral: \[ \int \frac{x^2 + 2x}{\sqrt{x^2 + 1}} \, dx \]

• A. \( \frac{1}{3} \left( x^2 + 1 \right)^{3/2} \)
• B. \( \frac{1}{2} \left( x^2 + 1 \right)^{3/2} \)
• C. \( \frac{1}{2} \left( x^2 + 1 \right)^{5/2} \)
• D. \( \frac{1}{3} \left( x^2 + 1 \right)^{5/2} \)

Hint:

Remember: For integrals involving square roots, substitution can simplify the process significantly.

Answer 1

The Correct Option is B

Step 1: Simplify the integral The given integral is: \[ \int \frac{x^2 + 2x}{\sqrt{x^2 + 1}} \, dx \] We can simplify this by separating terms: \[ = \int \frac{x^2}{\sqrt{x^2 + 1}} \, dx + \int \frac{2x}{\sqrt{x^2 + 1}} \, dx \] 

Step 2: Solve each integral The second integral can be solved easily by substitution, let: \[ u = x^2 + 1 \quad \Rightarrow \quad du = 2x \, dx \] This gives: \[ \int \frac{2x}{\sqrt{x^2 + 1}} \, dx = \int \frac{du}{\sqrt{u}} = 2\sqrt{u} = 2\sqrt{x^2 + 1} \] Now, the first integral can be simplified using substitution as well: \[ \int \frac{x^2}{\sqrt{x^2 + 1}} \, dx = \int \sqrt{x^2 + 1} \, dx \] This can be solved by standard methods for integrating square roots. 

Answer: The final result is: \[ \frac{1}{2} \left( x^2 + 1 \right)^{3/2} + C \]