Integrate X³ √x²+1

Let $u = x^2+1$.
Then, differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = 2x$$

This means $du = 2x \, dx$, or $x \, dx = \frac{1}{2} du$.
Also, from $u = x^2+1$, we can express $x^2$ as $x^2 = u-1$.

The original integral is $\int x^3 \sqrt{x^2+1} \, dx$.
We can rewrite $x^3$ as $x^2 \cdot x$.
So, the integral becomes $\int x^2 \sqrt{x^2+1} \cdot x \, dx$.
Substitute $x^2 = u-1$, $\sqrt{x^2+1} = \sqrt{u}$, and $x \, dx = \frac{1}{2} du$:

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

Distribute $u^{1/2}$:

$$\frac{1}{2} \int (u^{3/2} - u^{1/2}) du$$

Apply the power rule for integration, $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$:

$$\frac{1}{2} \left( \frac{u^{3/2+1}}{3/2+1} - \frac{u^{1/2+1}}{1/2+1} \right) + C$$ $$\frac{1}{2} \left( \frac{u^{5/2}}{5/2} - \frac{u^{3/2}}{3/2} \right) + C$$ $$\frac{1}{2} \left( \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right) + C$$ $$\frac{1}{5} u^{5/2} - \frac{1}{3} u^{3/2} + C$$

Factor out the common term $(x^2+1)^{3/2}$:

$$(x^2+1)^{3/2} \left( \frac{1}{5} (x^2+1) - \frac{1}{3} \right) + C$$

Find a common denominator for the terms inside the parenthesis:

$$(x^2+1)^{3/2} \left( \frac{3(x^2+1) - 5}{15} \right) + C$$ $$(x^2+1)^{3/2} \left( \frac{3x^2+3 - 5}{15} \right) + C$$ $$(x^2+1)^{3/2} \left( \frac{3x^2-2}{15} \right) + C$$ $$\frac{1}{15} (3x^2-2) (x^2+1)^{3/2} + C$$