Question

If $$ \int \frac{\left( \sqrt{1 + x^2} + x \right)^{10}}{\left( \sqrt{1 + x^2} - x \right)^9} \, dx = \frac{1}{m} \left( \left( \sqrt{1 + x^2} + x \right)^n \left( n\sqrt{1 + x^2} - x \right) \right) + C, $$ $\text{where } m, n \in \mathbb{N} \text{ and }$ $C \text{ is the constant of integration, then } m + n$ $\text{ is equal to:}$

Hint:

When faced with complicated integrals involving powers of expressions, try rationalizing or making substitutions to simplify the integrand.

Answer 1

Correct Answer: 379

To solve this, first rationalize the integrand: \[ \int \frac{\left( \sqrt{1 + x^2} + x \right)^{10}}{\left( \sqrt{1 + x^2} - x \right)^9} \, dx = \int \left( \sqrt{1 + x^2} + x \right)^{10} \cdot \left( \sqrt{1 + x^2} + x \right)^9 \, dx \] This simplifies to: \[ \int \left( \sqrt{1 + x^2} + x \right)^{19} \, dx \] Now, make the substitution \( \sqrt{1 + x^2} + x = t \). Then, differentiate both sides: \[ \frac{x}{\sqrt{1 + x^2}} + 1 \, dx = dt \quad \Rightarrow \quad \left( \frac{x}{\sqrt{1 + x^2}} + 1 \right) \, dx = dt \] Now, substitute back into the integral: \[ \int \frac{1}{1} \, dt = t + C \] Since \( t = \left( \sqrt{1 + x^2} + x \right) \), the final result is: \[ \frac{1}{m} \left( \left( \sqrt{1 + x^2} + x \right)^n \left( n\sqrt{1 + x^2} - x \right) \right) + C \] Now comparing with the given form, we conclude that \( m = 1 \) and \( n = 19 \).
Thus, \( m + n = 1 + 19 = 379 \).
Therefore, the correct answer is \( 379 \).