Solve ∫ 10x(6x+5)^2 wrt x

$$\int 10 x ( 6 x + 5 ) ^ { 2 } d x$$

Evaluate

$$90x^{4}+200x^{3}+125x^{2}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6x+5\right)^{2}.

$$\int 10x\left(36x^{2}+60x+25\right)\mathrm{d}x$$

Use the distributive property to multiply 10x by 36x^{2}+60x+25.

$$\int 360x^{3}+600x^{2}+250x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 360x^{3}\mathrm{d}x+\int 600x^{2}\mathrm{d}x+\int 250x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$360\int x^{3}\mathrm{d}x+600\int x^{2}\mathrm{d}x+250\int x\mathrm{d}x$$

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

$$90x^{4}+600\int x^{2}\mathrm{d}x+250\int x\mathrm{d}x$$

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

$$90x^{4}+200x^{3}+250\int x\mathrm{d}x$$

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

$$90x^{4}+200x^{3}+125x^{2}$$

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.

$$125x^{2}+200x^{3}+90x^{4}+С$$

Differentiate w.r.t. x

$$10x\left(6x+5\right)^{2}$$