Solve ∫ 3x(9x+6)^3 wrt x=?

$$\int 3 x ( 9 x + 6 ) ^ { 3 } d x = ?$$

Evaluate

$$\frac{2187x^{5}}{5}+\frac{2187x^{4}}{2}+972x^{3}+324x^{2}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(9x+6\right)^{3}.

$$\int 3x\left(729x^{3}+1458x^{2}+972x+216\right)\mathrm{d}x$$

Use the distributive property to multiply 3x by 729x^{3}+1458x^{2}+972x+216.

$$\int 2187x^{4}+4374x^{3}+2916x^{2}+648x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2187x^{4}\mathrm{d}x+\int 4374x^{3}\mathrm{d}x+\int 2916x^{2}\mathrm{d}x+\int 648x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2187\int x^{4}\mathrm{d}x+4374\int x^{3}\mathrm{d}x+2916\int x^{2}\mathrm{d}x+648\int x\mathrm{d}x$$

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 2187 times \frac{x^{5}}{5}.

$$\frac{2187x^{5}}{5}+4374\int x^{3}\mathrm{d}x+2916\int x^{2}\mathrm{d}x+648\int x\mathrm{d}x$$

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 4374 times \frac{x^{4}}{4}.

$$\frac{2187x^{5}}{5}+\frac{2187x^{4}}{2}+2916\int x^{2}\mathrm{d}x+648\int x\mathrm{d}x$$

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 2916 times \frac{x^{3}}{3}.

$$\frac{2187x^{5}}{5}+\frac{2187x^{4}}{2}+972x^{3}+648\int x\mathrm{d}x$$

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

$$\frac{2187x^{5}}{5}+\frac{2187x^{4}}{2}+972x^{3}+324x^{2}$$

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.

$$\frac{2187x^{5}}{5}+\frac{2187x^{4}}{2}+972x^{3}+324x^{2}+С$$

Differentiate w.r.t. x

$$81x\left(3x+2\right)^{3}$$