Solve ∫ sqrt{x^3}+2x+5 wrt x

$$\int \sqrt { x ^ { 3 } } + 2 x + 5 d x$$

Evaluate

$$\frac{2x^{\frac{5}{2}}}{5}+x^{2}+5x+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

Find the integral of 5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{\frac{5}{2}}}{5}+x^{2}+5x$$

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.

$$\frac{2x^{\frac{5}{2}}}{5}+x^{2}+5x+С$$

Differentiate w.r.t. x

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