Solve ∫ 2x(2x-3)^3 wrt x=

$$\int 2 x ( 2 x - 3 ) ^ { 3 } d x =$$

Evaluate

$$\frac{16x^{5}}{5}-18x^{4}+36x^{3}-27x^{2}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(2x-3\right)^{3}.

$$\int 2x\left(8x^{3}-36x^{2}+54x-27\right)\mathrm{d}x$$

Use the distributive property to multiply 2x by 8x^{3}-36x^{2}+54x-27.

$$\int 16x^{4}-72x^{3}+108x^{2}-54x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 16x^{4}\mathrm{d}x+\int -72x^{3}\mathrm{d}x+\int 108x^{2}\mathrm{d}x+\int -54x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$16\int x^{4}\mathrm{d}x-72\int x^{3}\mathrm{d}x+108\int x^{2}\mathrm{d}x-54\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 16 times \frac{x^{5}}{5}.

$$\frac{16x^{5}}{5}-72\int x^{3}\mathrm{d}x+108\int x^{2}\mathrm{d}x-54\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -72 times \frac{x^{4}}{4}.

$$\frac{16x^{5}}{5}-18x^{4}+108\int x^{2}\mathrm{d}x-54\int x\mathrm{d}x$$

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 108 times \frac{x^{3}}{3}.

$$\frac{16x^{5}}{5}-18x^{4}+36x^{3}-54\int x\mathrm{d}x$$

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 -54 times \frac{x^{2}}{2}.

$$\frac{16x^{5}}{5}-18x^{4}+36x^{3}-27x^{2}$$

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.

$$\frac{16x^{5}}{5}-18x^{4}+36x^{3}-27x^{2}+С$$

Differentiate w.r.t. x

$$2x\left(2x-3\right)^{3}$$