Solve ∫ 3^x-6x wrt x

$$\int 3 ^ { x } - 6 x d x$$

Evaluate

$$\frac{3^{x}}{\ln(3)}-3x^{2}+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

$$\int 3^{x}\mathrm{d}x-6\int x\mathrm{d}x$$

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

$$\frac{3^{x}}{\ln(3)}-6\int x\mathrm{d}x$$

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 -6 times \frac{x^{2}}{2}.

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

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.

$$\frac{3^{x}}{\ln(3)}-3x^{2}+С$$

Differentiate w.r.t. x

$$3^{x}-6x$$