Solve ∫ (2w+3)^2dw

$$\int ( 2 w + 3 ) ^ { 2 } d w$$

Evaluate

$$\frac{\left(2w+3\right)^{3}}{6}+С$$

Solution Steps

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

$$\int 4w^{2}+12w+9\mathrm{d}w$$

Integrate the sum term by term.

$$\int 4w^{2}\mathrm{d}w+\int 12w\mathrm{d}w+\int 9\mathrm{d}w$$

Factor out the constant in each of the terms.

$$4\int w^{2}\mathrm{d}w+12\int w\mathrm{d}w+\int 9\mathrm{d}w$$

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

$$\frac{4w^{3}}{3}+12\int w\mathrm{d}w+\int 9\mathrm{d}w$$

Since \int w^{k}\mathrm{d}w=\frac{w^{k+1}}{k+1} for k\neq -1, replace \int w\mathrm{d}w with \frac{w^{2}}{2}. Multiply 12 times \frac{w^{2}}{2}.

$$\frac{4w^{3}}{3}+6w^{2}+\int 9\mathrm{d}w$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}w=aw.

$$\frac{4w^{3}}{3}+6w^{2}+9w$$

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

$$\frac{4w^{3}}{3}+6w^{2}+9w+С$$

Differentiate w.r.t. w

$$\left(2w+3\right)^{2}$$