Solve ∫ (10+10x+10^x+10sqrt{x}) wrt x

$$\int ( 10 + 10 x + 10 ^ { x } + 10 \sqrt { x } ) d x$$

Evaluate

$$\log(e)\times 10^{x}+5x^{2}+\frac{20x^{\frac{3}{2}}}{3}+10x+С$$

Solution Steps

Integrate the sum term by term.

$$\int 10\mathrm{d}x+\int 10x\mathrm{d}x+\int 10^{x}\mathrm{d}x+\int 10\sqrt{x}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 10\mathrm{d}x+10\int x\mathrm{d}x+\int 10^{x}\mathrm{d}x+10\int \sqrt{x}\mathrm{d}x$$

Find the integral of 10 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$10x+10\int x\mathrm{d}x+\int 10^{x}\mathrm{d}x+10\int \sqrt{x}\mathrm{d}x$$

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

$$10x+5x^{2}+\int 10^{x}\mathrm{d}x+10\int \sqrt{x}\mathrm{d}x$$

Use \int a^{b}\mathrm{d}b=\frac{a^{b}}{\ln(a)} from the table of common integrals to obtain the result.

$$10x+5x^{2}+\frac{10^{x}}{\ln(10)}+10\int \sqrt{x}\mathrm{d}x$$

Rewrite \sqrt{x} as x^{\frac{1}{2}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. Simplify. Multiply 10 times \frac{2x^{\frac{3}{2}}}{3}.

$$10x+5x^{2}+\frac{10^{x}}{\ln(10)}+\frac{20x^{\frac{3}{2}}}{3}$$

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.

$$10x+5x^{2}+\frac{10^{x}}{\ln(10)}+\frac{20x^{\frac{3}{2}}}{3}+С$$

Differentiate w.r.t. x

$$10^{x}+10x+10\sqrt{x}+10$$