Solve ∫ (x^3+6x^2-3/2x+x/x^2) wrt x

$$\int{ ( { x }^{ 3 } +6 { x }^{ 2 } - \frac{ 3 }{ 2 } x+ \frac{ x }{ { x }^{ 2 } } ) }d x$$

Evaluate

$$\ln(|x|)+\frac{x^{4}}{4}+2x^{3}-\frac{3x^{2}}{4}+С$$

Solution Steps

Rewrite x^{2} as xx. Cancel out x in both numerator and denominator.

$$\int x^{3}+6x^{2}-\frac{3}{2}x+\frac{1}{x}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int 6x^{2}\mathrm{d}x+\int -\frac{3x}{2}\mathrm{d}x+\int \frac{1}{x}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x+6\int x^{2}\mathrm{d}x-\frac{3\int x\mathrm{d}x}{2}+\int \frac{1}{x}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{4}}{4}+6\int x^{2}\mathrm{d}x-\frac{3\int x\mathrm{d}x}{2}+\int \frac{1}{x}\mathrm{d}x$$

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}. Multiply 6 times \frac{x^{3}}{3}.

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

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 -\frac{3}{2} times \frac{x^{2}}{2}.

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

Use \int \frac{1}{x}\mathrm{d}x=\ln(|x|) from the table of common integrals to obtain the result.

$$\frac{x^{4}}{4}+2x^{3}-\frac{3x^{2}}{4}+\ln(|x|)$$

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.

$$\frac{x^{4}}{4}+2x^{3}-\frac{3x^{2}}{4}+\ln(|x|)+С$$

Differentiate w.r.t. x

$$x^{3}+6x^{2}-\frac{3x}{2}+\frac{1}{x}$$