Solve ∫ (9x^2+8x) wrt x

$$\int ( 9 x ^ { 2 } + 8 x ) d x$$

Evaluate

$$3x^{3}+4x^{2}+С$$

Solution Steps

Integrate the sum term by term.

$$\int 9x^{2}\mathrm{d}x+\int 8x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$9\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 9 times \frac{x^{3}}{3}.

$$3x^{3}+8\int x\mathrm{d}x$$

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

$$3x^{3}+4x^{2}$$

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.

$$3x^{3}+4x^{2}+С$$

Differentiate w.r.t. x

$$x\left(9x+8\right)$$