Solve ∫ (12x-8)(3x^2-4x+5) wrt x=

$$\int ( 12 x - 8 ) ( 3 x ^ { 2 } - 4 x + 5 ) d x =$$

Evaluate

$$9x^{4}-24x^{3}+46x^{2}-40x+С$$

Solution Steps

Use the distributive property to multiply 12x-8 by 3x^{2}-4x+5 and combine like terms.

$$\int 36x^{3}-72x^{2}+92x-40\mathrm{d}x$$

Integrate the sum term by term.

$$\int 36x^{3}\mathrm{d}x+\int -72x^{2}\mathrm{d}x+\int 92x\mathrm{d}x+\int -40\mathrm{d}x$$

Factor out the constant in each of the terms.

$$36\int x^{3}\mathrm{d}x-72\int x^{2}\mathrm{d}x+92\int x\mathrm{d}x+\int -40\mathrm{d}x$$

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 36 times \frac{x^{4}}{4}.

$$9x^{4}-72\int x^{2}\mathrm{d}x+92\int x\mathrm{d}x+\int -40\mathrm{d}x$$

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 -72 times \frac{x^{3}}{3}.

$$9x^{4}-24x^{3}+92\int x\mathrm{d}x+\int -40\mathrm{d}x$$

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

$$9x^{4}-24x^{3}+46x^{2}+\int -40\mathrm{d}x$$

Find the integral of -40 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$9x^{4}-24x^{3}+46x^{2}-40x$$

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.

$$9x^{4}-24x^{3}+46x^{2}-40x+С$$

Differentiate w.r.t. x

$$36x^{3}-72x^{2}+92x-40$$