Evaluate the Integral integral of (x^2+4x+1/(x^3)) with respect to x

$\int (x^{2} + 4 x + \frac{1}{x^{3}}) d x$

Remove parentheses.

$\int x^{2} + 4 x + \frac{1}{x^{3}} d x$

Split the single integral into multiple integrals.

$\int x^{2} d x + \int 4 x d x + \int \frac{1}{x^{3}} d x$

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$\frac{1}{3} x^{3} + C + \int 4 x d x + \int \frac{1}{x^{3}} d x$

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

$\frac{1}{3} x^{3} + C + 4 \int x d x + \int \frac{1}{x^{3}} d x$

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$\frac{1}{3} x^{3} + C + 4 (\frac{1}{2} x^{2} + C) + \int \frac{1}{x^{3}} d x$

Simplify the expression.

Move $x^{3}$ out of the denominator by raising it to the $- 1$ power.

$\frac{1}{3} x^{3} + C + 4 (\frac{1}{2} x^{2} + C) + \int {(x^{3})}^{- 1} d x$

Simplify.

Combine $\frac{1}{2}$ and $x^{2}$ .

$\frac{1}{3} x^{3} + C + 4 (\frac{x^{2}}{2} + C) + \int {(x^{3})}^{- 1} d x$

Multiply the exponents in ${(x^{3})}^{- 1}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{1}{3} x^{3} + C + 4 (\frac{x^{2}}{2} + C) + \int x^{3 \cdot - 1} d x$

Multiply $3$ by $- 1$ .

$\frac{1}{3} x^{3} + C + 4 (\frac{x^{2}}{2} + C) + \int x^{- 3} d x$

By the Power Rule, the integral of $x^{- 3}$ with respect to $x$ is $- \frac{1}{2} x^{- 2}$ .

$\frac{1}{3} x^{3} + C + 4 (\frac{x^{2}}{2} + C) - \frac{1}{2} x^{- 2} + C$

Simplify.

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

Reorder terms.

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