Solve ∫ (2x^3-15)^2 wrt x

$$\int ( 2 x ^ { 3 } - 15 ) ^ { 2 } d x$$

Evaluate

$$\frac{4x^{7}}{7}-15x^{4}+225x+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x^{3}-15\right)^{2}.

$$\int 4\left(x^{3}\right)^{2}-60x^{3}+225\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$\int 4x^{6}-60x^{3}+225\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{6}\mathrm{d}x+\int -60x^{3}\mathrm{d}x+\int 225\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{6}\mathrm{d}x-60\int x^{3}\mathrm{d}x+\int 225\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply 4 times \frac{x^{7}}{7}.

$$\frac{4x^{7}}{7}-60\int x^{3}\mathrm{d}x+\int 225\mathrm{d}x$$

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 -60 times \frac{x^{4}}{4}.

$$\frac{4x^{7}}{7}-15x^{4}+\int 225\mathrm{d}x$$

Find the integral of 225 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{4x^{7}}{7}-15x^{4}+225x$$

Simplify.

$$225x-15x^{4}+\frac{4x^{7}}{7}$$

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.

$$225x-15x^{4}+\frac{4x^{7}}{7}+С$$

Differentiate w.r.t. x

$$\left(2x^{3}-15\right)^{2}$$