What is the integral from 0 to 2 of (x^2+1)^2 ?

The integral from 0 to 2 of (x^2+1)^2 is 206/15

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integral from 0 to 2 of (x^2+1)^2

\int_{0}^{2} (x^{2} + 1)^{2} d x

\frac{206}{15}

Decimal Notation

13.73333 \dots

Solution steps

\int_{0}^{2} (x^{2} + 1)^{2} d x

Expand

(x^{2} + 1)^{2} : x^{4} + 2 x^{2} + 1

= \int_{0}^{2} x^{4} + 2 x^{2} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2} x^{4} d x + \int_{0}^{2} 2 x^{2} d x + \int_{0}^{2} 1 d x

\int_{0}^{2} x^{4} d x = \frac{32}{5}

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

\int_{0}^{2} 1 d x = 2

= \frac{32}{5} + \frac{16}{3} + 2

Simplify

\frac{32}{5} + \frac{16}{3} + 2 : \frac{206}{15}

= \frac{206}{15}

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What is the integral from 0 to 2 of (x^2+1)^2 ?
The integral from 0 to 2 of (x^2+1)^2 is 206/15