Solve ∫ 4(4x-3)^2 wrt x=

$$\int 4 ( 4 x - 3 ) ^ { 2 } d x =$$

Evaluate

$$\frac{\left(4x-3\right)^{3}}{3}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$\int 4\left(16x^{2}-24x+9\right)\mathrm{d}x$$

Use the distributive property to multiply 4 by 16x^{2}-24x+9.

$$\int 64x^{2}-96x+36\mathrm{d}x$$

Integrate the sum term by term.

$$\int 64x^{2}\mathrm{d}x+\int -96x\mathrm{d}x+\int 36\mathrm{d}x$$

Factor out the constant in each of the terms.

$$64\int x^{2}\mathrm{d}x-96\int x\mathrm{d}x+\int 36\mathrm{d}x$$

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

$$\frac{64x^{3}}{3}-96\int x\mathrm{d}x+\int 36\mathrm{d}x$$

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 -96 times \frac{x^{2}}{2}.

$$\frac{64x^{3}}{3}-48x^{2}+\int 36\mathrm{d}x$$

Find the integral of 36 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{64x^{3}}{3}-48x^{2}+36x$$

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.

$$\frac{64x^{3}}{3}-48x^{2}+36x+С$$

Differentiate w.r.t. x

$$4\left(4x-3\right)^{2}$$