Solve ∫ (x^3-1)^2 wrt x=

$$\int ( x ^ { 3 } - 1 ) ^ { 2 } d x =$$

Evaluate

$$\frac{x^{7}}{7}-\frac{x^{4}}{2}+x+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{3}-1\right)^{2}.

$$\int \left(x^{3}\right)^{2}-2x^{3}+1\mathrm{d}x$$

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

$$\int x^{6}-2x^{3}+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{6}\mathrm{d}x+\int -2x^{3}\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{6}\mathrm{d}x-2\int x^{3}\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

$$\frac{x^{7}}{7}-2\int x^{3}\mathrm{d}x+\int 1\mathrm{d}x$$

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 -2 times \frac{x^{4}}{4}.

$$\frac{x^{7}}{7}-\frac{x^{4}}{2}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{7}}{7}-\frac{x^{4}}{2}+x$$

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.

$$x-\frac{x^{4}}{2}+\frac{x^{7}}{7}+С$$

Differentiate w.r.t. x

$$\left(x^{3}-1\right)^{2}$$