Solve ∫ 10x(5x+2)^2 wrt x=?

$$\int 10 x ( 5 x + 2 ) ^ { 2 } d x = ?$$

Evaluate

$$\frac{125x^{4}}{2}+\frac{200x^{3}}{3}+20x^{2}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+2\right)^{2}.

$$\int 10x\left(25x^{2}+20x+4\right)\mathrm{d}x$$

Use the distributive property to multiply 10x by 25x^{2}+20x+4.

$$\int 250x^{3}+200x^{2}+40x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 250x^{3}\mathrm{d}x+\int 200x^{2}\mathrm{d}x+\int 40x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$250\int x^{3}\mathrm{d}x+200\int x^{2}\mathrm{d}x+40\int x\mathrm{d}x$$

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 250 times \frac{x^{4}}{4}.

$$\frac{125x^{4}}{2}+200\int x^{2}\mathrm{d}x+40\int x\mathrm{d}x$$

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

$$\frac{125x^{4}}{2}+\frac{200x^{3}}{3}+40\int x\mathrm{d}x$$

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 40 times \frac{x^{2}}{2}.

$$\frac{125x^{4}}{2}+\frac{200x^{3}}{3}+20x^{2}$$

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.

$$\frac{125x^{4}}{2}+\frac{200x^{3}}{3}+20x^{2}+С$$

Differentiate w.r.t. x

$$10x\left(5x+2\right)^{2}$$