Question

$\frac{x^3}{\sqrt{x^2 + a^2}}$

Answer 1

$\frac{1}{3} (x^2 - 2a^2) \sqrt{x^2 + a^2} + C$

Answer 2

To evaluate the integral $\int \frac{x^3}{\sqrt{x^2 + a^2}} dx$, we use the method of substitution.

1. Substitution:
   Let $u = x^2 + a^2$.
   Differentiating both sides with respect to $x$, we get $du = 2x \, dx$.
   From this, we have $x \, dx = \frac{1}{2} du$.
   Also, from the substitution, $x^2 = u - a^2$.

2. Rewrite the Integral:
   The integral can be written as $\int \frac{x^2 \cdot x}{\sqrt{x^2 + a^2}} dx$.
   Substitute $x^2 = u - a^2$, $x \, dx = \frac{1}{2} du$, and $\sqrt{x^2 + a^2} = \sqrt{u}$: $\int \frac{(u - a^2)}{\sqrt{u}} \cdot \frac{1}{2} du$

3. Simplify and Integrate: $\frac{1}{2} \int \left( \frac{u}{\sqrt{u}} - \frac{a^2}{\sqrt{u}} \right) du = \frac{1}{2} \int (u^{1/2} - a^2 u^{-1/2}) du$ Now, integrate term by term: $\frac{1}{2} \left( \frac{u^{1/2+1}}{1/2+1} - a^2 \frac{u^{-1/2+1}}{-1/2+1} \right) + C$ $\frac{1}{2} \left( \frac{u^{3/2}}{3/2} - a^2 \frac{u^{1/2}}{1/2} \right) + C$ $\frac{1}{2} \left( \frac{2}{3} u^{3/2} - 2a^2 u^{1/2} \right) + C$ $\frac{1}{3} u^{3/2} - a^2 u^{1/2} + C$

4. Substitute Back:
   Replace $u$ with $x^2 + a^2$: $\frac{1}{3} (x^2 + a^2)^{3/2} - a^2 (x^2 + a^2)^{1/2} + C$

5. Factor and Simplify:
   Factor out $(x^2 + a^2)^{1/2}$: $(x^2 + a^2)^{1/2} \left[ \frac{1}{3} (x^2 + a^2) - a^2 \right] + C$ $\sqrt{x^2 + a^2} \left[ \frac{x^2 + a^2 - 3a^2}{3} \right] + C$ $\sqrt{x^2 + a^2} \left[ \frac{x^2 - 2a^2}{3} \right] + C$ $\frac{1}{3} (x^2 - 2a^2) \sqrt{x^2 + a^2} + C$