Solve ∫ 5x(3x+6)^3 wrt x=?

$$\int 5 x ( 3 x + 6 ) ^ { 3 } d x = ?$$

Evaluate

$$27x^{5}+\frac{405x^{4}}{2}+540x^{3}+540x^{2}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(3x+6\right)^{3}.

$$\int 5x\left(27x^{3}+162x^{2}+324x+216\right)\mathrm{d}x$$

Use the distributive property to multiply 5x by 27x^{3}+162x^{2}+324x+216.

$$\int 135x^{4}+810x^{3}+1620x^{2}+1080x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 135x^{4}\mathrm{d}x+\int 810x^{3}\mathrm{d}x+\int 1620x^{2}\mathrm{d}x+\int 1080x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$135\int x^{4}\mathrm{d}x+810\int x^{3}\mathrm{d}x+1620\int x^{2}\mathrm{d}x+1080\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 135 times \frac{x^{5}}{5}.

$$27x^{5}+810\int x^{3}\mathrm{d}x+1620\int x^{2}\mathrm{d}x+1080\int x\mathrm{d}x$$

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

$$27x^{5}+\frac{405x^{4}}{2}+1620\int x^{2}\mathrm{d}x+1080\int x\mathrm{d}x$$

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 1620 times \frac{x^{3}}{3}.

$$27x^{5}+\frac{405x^{4}}{2}+540x^{3}+1080\int x\mathrm{d}x$$

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 1080 times \frac{x^{2}}{2}.

$$27x^{5}+\frac{405x^{4}}{2}+540x^{3}+540x^{2}$$

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.

$$540x^{2}+540x^{3}+\frac{405x^{4}}{2}+27x^{5}+С$$

Differentiate w.r.t. x

$$135x\left(x+2\right)^{3}$$