Solve ∫ (from 0 to 1) of x(x^2+1)^3

$$\int _ { 0 } ^ { 1 } x ( x ^ { 2 } + 1 ) ^ { 3 }$$

Evaluate

$$\frac{15}{8}=1.875$$

Solution Steps

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

$$\int _{0}^{1}x\left(\left(x^{2}\right)^{3}+3\left(x^{2}\right)^{2}+3x^{2}+1\right)\mathrm{d}x$$

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

$$\int _{0}^{1}x\left(x^{6}+3\left(x^{2}\right)^{2}+3x^{2}+1\right)\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int _{0}^{1}x\left(x^{6}+3x^{4}+3x^{2}+1\right)\mathrm{d}x$$

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

$$\int _{0}^{1}x^{7}+3x^{5}+3x^{3}+x\mathrm{d}x$$

Evaluate the indefinite integral first.

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

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}.

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

Simplify.

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

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{1^{2}}{2}+\frac{3}{4}\times 1^{4}+\frac{1^{6}}{2}+\frac{1^{8}}{8}-\left(\frac{0^{2}}{2}+\frac{3}{4}\times 0^{4}+\frac{0^{6}}{2}+\frac{0^{8}}{8}\right)$$

Simplify.

$$\frac{15}{8}$$