Evaluate integral (x² + 1) \,dx

We use the power rule for integration, $\int x^n \,dx = \frac{x^{n+1}}{n+1} + C$.

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

The integral is $\boxed{\frac{x^3}{3} + x + C}$.

First, rewrite the integrand using exponent notation: $\sqrt[3]{x^2} = x^{2/3}$.
Then, $\frac{1}{x^{2/3}} = x^{-2/3}$.

$$\int x^{-2/3} \,dx$$

Apply the power rule for integration:

$$= \frac{x^{-2/3 + 1}}{-2/3 + 1} + C$$ $$= \frac{x^{1/3}}{1/3} + C$$ $$= 3x^{1/3} + C$$ $$= 3\sqrt[3]{x} + C$$

The integral is $\boxed{3\sqrt[3]{x} + C}$.

Integrate each term using the power rule:

$$\int (3x^2 + 2x + 5) \,dx = \int 3x^2 \,dx + \int 2x \,dx + \int 5 \,dx$$ $$= 3\frac{x^{2+1}}{2+1} + 2\frac{x^{1+1}}{1+1} + 5x + C$$ $$= 3\frac{x^3}{3} + 2\frac{x^2}{2} + 5x + C$$ $$= x^3 + x^2 + 5x + C$$

The integral is $\boxed{x^3 + x^2 + 5x + C}$.

First, divide each term in the numerator by $x^2$:

$$\frac{x^4}{x^2} + \frac{3x^3}{x^2} - \frac{4x}{x^2} = x^{4-2} + 3x^{3-2} - 4x^{1-2} = x^2 + 3x - 4x^{-1}$$

Now integrate the simplified expression:

$$\int (x^2 + 3x - 4x^{-1}) \,dx = \int x^2 \,dx + \int 3x \,dx - \int 4x^{-1} \,dx$$ $$= \frac{x^{2+1}}{2+1} + 3\frac{x^{1+1}}{1+1} - 4\ln|x| + C$$ $$= \frac{x^3}{3} + 3\frac{x^2}{2} - 4\ln|x| + C$$

The integral is $\boxed{\frac{x^3}{3} + \frac{3x^2}{2} - 4\ln|x| + C}$.

First, expand the integrand:

$$x^2 (3x^2 + 4x) = 3x^2 \cdot x^2 + 4x \cdot x^2 = 3x^4 + 4x^3$$

Now integrate the expanded expression:

$$\int (3x^4 + 4x^3) \,dx = \int 3x^4 \,dx + \int 4x^3 \,dx$$ $$= 3\frac{x^{4+1}}{4+1} + 4\frac{x^{3+1}}{3+1} + C$$ $$= 3\frac{x^5}{5} + 4\frac{x^4}{4} + C$$ $$= \frac{3x^5}{5} + x^4 + C$$

The integral is $\boxed{\frac{3x^5}{5} + x^4 + C}$.

First, divide each term in the numerator by $x$:

$$\frac{x^2}{x} - \frac{1}{x} = x - \frac{1}{x}$$

Now integrate the simplified expression:

$$\int \left(x - \frac{1}{x}\right) \,dx = \int x \,dx - \int \frac{1}{x} \,dx$$ $$= \frac{x^{1+1}}{1+1} - \ln|x| + C$$ $$= \frac{x^2}{2} - \ln|x| + C$$

The integral is $\boxed{\frac{x^2}{2} - \ln|x| + C}$.

First, expand the product of the three binomials:
Multiply the first two terms:

$$(x-2)(2x+3) = x(2x) + x(3) - 2(2x) - 2(3) = 2x^2 + 3x - 4x - 6 = 2x^2 - x - 6$$

Now multiply the result by $(x-5)$:

$$(2x^2 - x - 6)(x-5) = 2x^2(x-5) - x(x-5) - 6(x-5)$$ $$= (2x^3 - 10x^2) - (x^2 - 5x) - (6x - 30)$$ $$= 2x^3 - 10x^2 - x^2 + 5x - 6x + 30$$ $$= 2x^3 - 11x^2 - x + 30$$

Now integrate the expanded polynomial:

$$\int (2x^3 - 11x^2 - x + 30) \,dx$$ $$= \int 2x^3 \,dx - \int 11x^2 \,dx - \int x \,dx + \int 30 \,dx$$ $$= 2\frac{x^{3+1}}{3+1} - 11\frac{x^{2+1}}{2+1} - \frac{x^{1+1}}{1+1} + 30x + C$$ $$= 2\frac{x^4}{4} - 11\frac{x^3}{3} - \frac{x^2}{2} + 30x + C$$ $$= \frac{x^4}{2} - \frac{11x^3}{3} - \frac{x^2}{2} + 30x + C$$

The integral is $\boxed{\frac{x^4}{2} - \frac{11x^3}{3} - \frac{x^2}{2} + 30x + C}$.