Calculate the integral: integral from 0 to 1

Calculate the integral: $\int_0^1 \frac{x^3-2x+3}{\sqrt{x^2+1}} \, dx$.

We will split the integral and use substitution to evaluate each part.

Step 1: Split the integral

\[\int_0^1 \frac{x^3-2x+3}{\sqrt{x^2+1}} \, dx = \int_0^1 \frac{x^3}{\sqrt{x^2+1}} \, dx - \int_0^1 \frac{2x}{\sqrt{x^2+1}} \, dx + \int_0^1 \frac{3}{\sqrt{x^2+1}} \, dx\]

Step 2: Evaluate $\int_0^1 \frac{2x}{\sqrt{x^2+1}} \, dx$

Let $u = x^2+1$, $du=2x\,dx$. When $x=0$: $u=1$; when $x=1$: $u=2$.

\[\int_0^1 \frac{2x}{\sqrt{x^2+1}} \, dx = \int_1^2 u^{-1/2} \, du = \left[2u^{1/2}\right]_1^2 = 2\sqrt{2} - 2\]

Step 3: Evaluate $\int_0^1 \frac{3}{\sqrt{x^2+1}} \, dx$

We know that $\int \frac{1}{\sqrt{x^2+1}} \, dx = \ln(x+\sqrt{x^2+1}) + C$. Therefore:

\[\int_0^1 \frac{3}{\sqrt{x^2+1}} \, dx = 3\left[\ln(x + \sqrt{x^2+1})\right]_0^1 = 3[\ln(1+\sqrt2) - \ln 1] = 3\ln(1+\sqrt2)\]

Step 4: Evaluate $\int_0^1 \frac{x^3}{\sqrt{x^2+1}} \, dx$

Using $x^2=(x^2+1)-1$:

\[\frac{x^3}{\sqrt{x^2+1}} = x\left(\sqrt{x^2+1} - \frac{1}{\sqrt{x^2+1}}\right) = x\sqrt{x^2+1} - \frac{x}{\sqrt{x^2+1}}\]

Therefore:

\[\int_0^1 \frac{x^3}{\sqrt{x^2+1}} \, dx = \int_0^1 x\sqrt{x^2+1} \, dx - \int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx\]

For $\int_0^1 x\sqrt{x^2+1} \, dx$, let $u=x^2+1$:

\[\int_0^1 x\sqrt{x^2+1} \, dx = \frac{1}{2}\int_1^2 \sqrt{u} \, du = \frac{1}{3}\left[u^{3/2}\right]_1^2 = \frac{1}{3}(2\sqrt{2}-1)\]

For $\int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx$, let $u=x^2+1$:

\[\int_0^1 \frac{x}{\sqrt{x^2+1}} \, dx = \frac{1}{2}\int_1^2 u^{-1/2} \, du = \left[\sqrt{u}\right]_1^2 = \sqrt{2}-1\]

Therefore:

\[\int_0^1 \frac{x^3}{\sqrt{x^2+1}} \, dx = \frac{1}{3}(2\sqrt{2}-1) - (\sqrt{2}-1) = \frac{2-\sqrt{2}}{3}\]

Step 5: Combine all parts

\[\int_0^1 \frac{x^3-2x+3}{\sqrt{x^2+1}} \, dx = \frac{2-\sqrt{2}}{3} - (2\sqrt{2} - 2) + 3\ln(1 + \sqrt{2})\]

\[= \frac{2-\sqrt{2} + 6 - 6\sqrt{2}}{3} + 3\ln(1 + \sqrt{2}) = \frac{8 - 7\sqrt{2}}{3} + 3\ln(1 + \sqrt{2})\]

Numerical verification:

Using $\sqrt{2} \approx 1.4142$ and $\ln(1 + \sqrt{2}) \approx 0.8814$:

\[\frac{8 - 7(1.4142)}{3} + 3(0.8814) \approx -0.6331 + 2.6442 \approx 2.0111\]

Final answer:

\[\boxed{\int_0^1 \frac{x^3-2x+3}{\sqrt{x^2+1}} \, dx = \frac{8 - 7\sqrt{2}}{3} + 3\ln(1 + \sqrt{2})}\]

Or approximately:

\[\boxed{\int_0^1 \frac{x^3-2x+3}{\sqrt{x^2+1}} \, dx \approx 2.011}\]