Solve ∫ (4x-7)^2 wrt x

$$\int ( 4 x - 7 ) ^ { 2 } d x$$

Evaluate

$$\frac{\left(4x-7\right)^{3}}{12}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-7\right)^{2}.

$$\int 16x^{2}-56x+49\mathrm{d}x$$

Integrate the sum term by term.

$$\int 16x^{2}\mathrm{d}x+\int -56x\mathrm{d}x+\int 49\mathrm{d}x$$

Factor out the constant in each of the terms.

$$16\int x^{2}\mathrm{d}x-56\int x\mathrm{d}x+\int 49\mathrm{d}x$$

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

$$\frac{16x^{3}}{3}-56\int x\mathrm{d}x+\int 49\mathrm{d}x$$

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 -56 times \frac{x^{2}}{2}.

$$\frac{16x^{3}}{3}-28x^{2}+\int 49\mathrm{d}x$$

Find the integral of 49 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{16x^{3}}{3}-28x^{2}+49x$$

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.

$$\frac{16x^{3}}{3}-28x^{2}+49x+С$$

Differentiate w.r.t. x

$$\left(4x-7\right)^{2}$$