Solve ∫ x^3+3x^2+4x/x wrt x

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

Evaluate

$$\frac{x^{3}}{3}+\frac{3x^{2}}{2}+4x+С$$

Solution Steps

Factor the expressions that are not already factored in \frac{x^{3}+3x^{2}+4x}{x}.

$$\int \frac{x\left(x^{2}+3x+4\right)}{x}\mathrm{d}x$$

Cancel out x in both numerator and denominator.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 3 times \frac{x^{2}}{2}.

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

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}+\frac{3x^{2}}{2}+4x$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{x^{3}}{3}+\frac{3x^{2}}{2}+4x+С$$

Differentiate w.r.t. x

$$x^{2}+3x+4$$