Solve ∫ (x-1)^3(x^3+1)

$$\int ( x - 1 ) ^ { 3 } ( x ^ { 3 } + 1 )$$

Evaluate

$$\frac{x^{7}}{7}-\frac{x^{6}}{2}+\frac{3x^{5}}{5}-x^{3}+\frac{3x^{2}}{2}-x+С$$

Solution Steps

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

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

Use the distributive property to multiply x^{3}-3x^{2}+3x-1 by x^{3}+1 and combine like terms.

$$\int x^{6}-3x^{5}-3x^{2}+3x^{4}+3x-1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{6}\mathrm{d}x+\int -3x^{5}\mathrm{d}x+\int -3x^{2}\mathrm{d}x+\int 3x^{4}\mathrm{d}x+\int 3x\mathrm{d}x+\int -1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{6}\mathrm{d}x-3\int x^{5}\mathrm{d}x-3\int x^{2}\mathrm{d}x+3\int x^{4}\mathrm{d}x+3\int x\mathrm{d}x+\int -1\mathrm{d}x$$

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}.

$$\frac{x^{7}}{7}-3\int x^{5}\mathrm{d}x-3\int x^{2}\mathrm{d}x+3\int x^{4}\mathrm{d}x+3\int x\mathrm{d}x+\int -1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply -3 times \frac{x^{6}}{6}.

$$\frac{x^{7}}{7}-\frac{x^{6}}{2}-3\int x^{2}\mathrm{d}x+3\int x^{4}\mathrm{d}x+3\int x\mathrm{d}x+\int -1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -3 times \frac{x^{3}}{3}.

$$\frac{x^{7}}{7}-\frac{x^{6}}{2}-x^{3}+3\int x^{4}\mathrm{d}x+3\int x\mathrm{d}x+\int -1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 3 times \frac{x^{5}}{5}.

$$\frac{x^{7}}{7}-\frac{x^{6}}{2}-x^{3}+\frac{3x^{5}}{5}+3\int x\mathrm{d}x+\int -1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 3 times \frac{x^{2}}{2}.

$$\frac{x^{7}}{7}-\frac{x^{6}}{2}-x^{3}+\frac{3x^{5}}{5}+\frac{3x^{2}}{2}+\int -1\mathrm{d}x$$

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

$$\frac{x^{7}}{7}-\frac{x^{6}}{2}-x^{3}+\frac{3x^{5}}{5}+\frac{3x^{2}}{2}-x$$

Simplify.

$$-x+\frac{3x^{5}}{5}+\frac{x^{7}}{7}+\frac{3x^{2}}{2}-\frac{x^{6}}{2}-x^{3}$$

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.

$$-x+\frac{3x^{5}}{5}+\frac{x^{7}}{7}+\frac{3x^{2}}{2}-\frac{x^{6}}{2}-x^{3}+С$$

Differentiate w.r.t. x

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