Solve ∫ (x-1)^3(x+1)^3x wrt x

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

Evaluate

$$\frac{x^{8}}{8}-\frac{x^{6}}{2}+\frac{3x^{4}}{4}-\frac{x^{2}}{2}+С$$

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}.

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

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

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

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

$$\int \left(x^{6}-3x^{4}+3x^{2}-1\right)x\mathrm{d}x$$

Use the distributive property to multiply x^{6}-3x^{4}+3x^{2}-1 by x.

$$\int x^{7}-3x^{5}+3x^{3}-x\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{7}\mathrm{d}x+\int -3x^{5}\mathrm{d}x+\int 3x^{3}\mathrm{d}x+\int -x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{7}\mathrm{d}x-3\int x^{5}\mathrm{d}x+3\int x^{3}\mathrm{d}x-\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{7}\mathrm{d}x with \frac{x^{8}}{8}.

$$\frac{x^{8}}{8}-3\int x^{5}\mathrm{d}x+3\int x^{3}\mathrm{d}x-\int x\mathrm{d}x$$

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

$$\frac{x^{8}}{8}-\frac{x^{6}}{2}+3\int x^{3}\mathrm{d}x-\int x\mathrm{d}x$$

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

$$\frac{x^{8}}{8}-\frac{x^{6}}{2}+\frac{3x^{4}}{4}-\int x\mathrm{d}x$$

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 -1 times \frac{x^{2}}{2}.

$$\frac{x^{8}}{8}-\frac{x^{6}}{2}+\frac{3x^{4}}{4}-\frac{x^{2}}{2}$$

Simplify.

$$-\frac{x^{2}}{2}+\frac{3x^{4}}{4}-\frac{x^{6}}{2}+\frac{x^{8}}{8}$$

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.

$$-\frac{x^{2}}{2}+\frac{3x^{4}}{4}-\frac{x^{6}}{2}+\frac{x^{8}}{8}+С$$

Differentiate w.r.t. x

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