Solve ∫ (1-2u)^2du

$$\int ( 1 - 2 u ) ^ { 2 } d u$$

Evaluate

$$-\frac{\left(1-2u\right)^{3}}{6}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-2u\right)^{2}.

$$\int 1-4u+4u^{2}\mathrm{d}u$$

Integrate the sum term by term.

$$\int 1\mathrm{d}u+\int -4u\mathrm{d}u+\int 4u^{2}\mathrm{d}u$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}u-4\int u\mathrm{d}u+4\int u^{2}\mathrm{d}u$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}u=au.

$$u-4\int u\mathrm{d}u+4\int u^{2}\mathrm{d}u$$

Since \int u^{k}\mathrm{d}u=\frac{u^{k+1}}{k+1} for k\neq -1, replace \int u\mathrm{d}u with \frac{u^{2}}{2}. Multiply -4 times \frac{u^{2}}{2}.

$$u-2u^{2}+4\int u^{2}\mathrm{d}u$$

Since \int u^{k}\mathrm{d}u=\frac{u^{k+1}}{k+1} for k\neq -1, replace \int u^{2}\mathrm{d}u with \frac{u^{3}}{3}. Multiply 4 times \frac{u^{3}}{3}.

$$u-2u^{2}+\frac{4u^{3}}{3}$$

If F\left(u\right) is an antiderivative of f\left(u\right), then the set of all antiderivatives of f\left(u\right) is given by F\left(u\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$u-2u^{2}+\frac{4u^{3}}{3}+С$$

Differentiate w.r.t. u

$$\left(1-2u\right)^{2}$$