What is the integral from 0 to 2 of 7x(x^2+1)^2 ?

The integral from 0 to 2 of 7x(x^2+1)^2 is 434/3

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integral from 0 to 2 of 7x(x^2+1)^2

\int_{0}^{2} 7 x (x^{2} + 1)^{2} d x

\frac{434}{3}

Decimal Notation

144.66666 \dots

Solution steps

\int_{0}^{2} 7 x (x^{2} + 1)^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int_{0}^{2} x (x^{2} + 1)^{2} d x

Expand

x (x^{2} + 1)^{2} : x^{5} + 2 x^{3} + x

= 7 \cdot \int_{0}^{2} x^{5} + 2 x^{3} + x 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

= 7 (\int_{0}^{2} x^{5} d x + \int_{0}^{2} 2 x^{3} d x + \int_{0}^{2} x d x)

\int_{0}^{2} x^{5} d x = \frac{32}{3}

\int_{0}^{2} 2 x^{3} d x = 8

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

= 7 (\frac{32}{3} + 8 + 2)

Simplify

7 (\frac{32}{3} + 8 + 2) : \frac{434}{3}

= \frac{434}{3}

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What is the integral from 0 to 2 of 7x(x^2+1)^2 ?
The integral from 0 to 2 of 7x(x^2+1)^2 is 434/3