What is the integral from 0 to 2 of 17x(x^2+1)^2 ?

The integral from 0 to 2 of 17x(x^2+1)^2 is 1054/3

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integral from 0 to 2 of 17x(x^2+1)^2

\int_{0}^{2} 17 x (x^{2} + 1)^{2} d x

\frac{1054}{3}

Decimal

351.33333 \dots

Solution steps

\int_{0}^{2} 17 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

= 17 \cdot \int_{0}^{2} x (x^{2} + 1)^{2} d x

Expand

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

= 17 \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

= 17 (\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

= 17 (\frac{32}{3} + 8 + 2)

Simplify

17 (\frac{32}{3} + 8 + 2) : \frac{1054}{3}

= \frac{1054}{3}

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

\int_{- 5}^{5} (7 + 25 - x^{2}) d x

\int_{1}^{8} x^{\frac{2}{3}} d x

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

\int_{1}^{2} x^{2} - x^{- 2} d x

What is the integral from 0 to 2 of 17x(x^2+1)^2 ?
The integral from 0 to 2 of 17x(x^2+1)^2 is 1054/3