Question

$\int \frac{x^3dx}{\sqrt{1+x^2}}$

Answer 1

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

Answer 2

To evaluate the indefinite integral $\int \frac{x^3dx}{\sqrt{1+x^2}}$, we can use substitution.

1. Substitution: Let $u = 1+x^2$. Then $du = 2x dx$, and $x^2 = u-1$. Rewrite the integrand: $\frac{x^3}{\sqrt{1+x^2}} dx = \frac{x^2 \cdot x dx}{\sqrt{1+x^2}}$.

2. Transform the integral: Substitute the expressions in terms of $u$:

   $$\int \frac{(u-1)}{\sqrt{u}} \left(\frac{1}{2} du\right) = \frac{1}{2} \int \left(\frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}}\right) du = \frac{1}{2} \int (u^{1/2} - u^{-1/2}) du$$

3. Integrate with respect to $u$: Use the power rule for integration $\int t^n dt = \frac{t^{n+1}}{n+1} + C$.

   $$\frac{1}{2} \left( \frac{u^{1/2+1}}{1/2+1} - \frac{u^{-1/2+1}}{-1/2+1} \right) + C = \frac{1}{2} \left( \frac{u^{3/2}}{3/2} - \frac{u^{1/2}}{1/2} \right) + C = \frac{1}{2} \left( \frac{2}{3} u^{3/2} - 2 u^{1/2} \right) + C = \frac{1}{3} u^{3/2} - u^{1/2} + C$$

4. Substitute back $u = 1+x^2$:

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

5. Simplify the expression: Factor out $(1+x^2)^{1/2}$:

   $$(1+x^2)^{1/2} \left( \frac{1}{3} (1+x^2) - 1 \right) + C = \sqrt{1+x^2} \left( \frac{1+x^2-3}{3} \right) + C = \sqrt{1+x^2} \left( \frac{x^2-2}{3} \right) + C = \frac{(x^2-2)\sqrt{1+x^2}}{3} + C$$

Thus, the integral is:

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