Solve ∫ (2x^3-3sqrt{x^5}+4/x) wrt x=

$$\int{ (2 { x }^{ 3 } -3 \sqrt{ { x }^{ 5 } } + \frac{ 4 }{ x } ) }d x =$$

Evaluate

$$4\ln(x)+\frac{x^{4}}{2}-\frac{6x^{\frac{7}{2}}}{7}+С$$

Solution Steps

Integrate the sum term by term.

$$\int 2x^{3}\mathrm{d}x+\int -3x^{\frac{5}{2}}\mathrm{d}x+\int \frac{4}{x}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{3}\mathrm{d}x-3\int x^{\frac{5}{2}}\mathrm{d}x+4\int \frac{1}{x}\mathrm{d}x$$

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}. Multiply 2 times \frac{x^{4}}{4}.

$$\frac{x^{4}}{2}-3\int x^{\frac{5}{2}}\mathrm{d}x+4\int \frac{1}{x}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{5}{2}}\mathrm{d}x with \frac{2x^{\frac{7}{2}}}{7}. Multiply -3 times \frac{2x^{\frac{7}{2}}}{7}.

$$\frac{x^{4}}{2}-\frac{6x^{\frac{7}{2}}}{7}+4\int \frac{1}{x}\mathrm{d}x$$

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

$$\frac{x^{4}}{2}-\frac{6x^{\frac{7}{2}}}{7}+4\ln(|x|)$$

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.

$$\frac{x^{4}}{2}-\frac{6x^{\frac{7}{2}}}{7}+4\ln(x)+С$$

Differentiate w.r.t. x

$$2x^{3}-3x^{\frac{5}{2}}+\frac{4}{x}$$