What is the integral from 1 to 3 of (x^2+1)^3 ?

The integral from 1 to 3 of (x^2+1)^3 is 16992/35

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integral from 1 to 3 of (x^2+1)^3

\int_{1}^{3} (x^{2} + 1)^{3} d x

\frac{16992}{35}

Decimal

485.48571 \dots

Solution steps

\int_{1}^{3} (x^{2} + 1)^{3} d x

Expand

(x^{2} + 1)^{3} : x^{6} + 3 x^{4} + 3 x^{2} + 1

= \int_{1}^{3} x^{6} + 3 x^{4} + 3 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_{1}^{3} x^{6} d x + \int_{1}^{3} 3 x^{4} d x + \int_{1}^{3} 3 x^{2} d x + \int_{1}^{3} 1 d x

\int_{1}^{3} x^{6} d x = \frac{2186}{7}

\int_{1}^{3} 3 x^{4} d x = \frac{726}{5}

\int_{1}^{3} 3 x^{2} d x = 26

\int_{1}^{3} 1 d x = 2

= \frac{2186}{7} + \frac{726}{5} + 26 + 2

Simplify

\frac{2186}{7} + \frac{726}{5} + 26 + 2 : \frac{16992}{35}

= \frac{16992}{35}

What is the integral from 1 to 3 of (x^2+1)^3 ?
The integral from 1 to 3 of (x^2+1)^3 is 16992/35