Solve ∫ (from 0 to 2) of (1+x^2)^2 wrt x

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

Evaluate

$$\frac{206}{15}\approx 13.733333333$$

Solution Steps

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

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

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

$$\int _{0}^{2}1+2x^{2}+x^{4}\mathrm{d}x$$

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

$$x+2\int x^{2}\mathrm{d}x+\int x^{4}\mathrm{d}x$$

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

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

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}.

$$x+\frac{2x^{3}}{3}+\frac{x^{5}}{5}$$

Simplify.

$$\frac{x^{5}}{5}+\frac{2x^{3}}{3}+x$$

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.

$$\frac{2^{5}}{5}+\frac{2}{3}\times 2^{3}+2-\left(\frac{0^{5}}{5}+\frac{2}{3}\times 0^{3}+0\right)$$

Simplify.

$$\frac{206}{15}$$