What is the integral from 0 to 1 of 6x(x^2+1)^2 ?

The integral from 0 to 1 of 6x(x^2+1)^2 is 7

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integral from 0 to 1 of 6x(x^2+1)^2

\int_{0}^{1} 6 x (x^{2} + 1)^{2} d x

7

Solution steps

\int_{0}^{1} 6 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

= 6 \cdot \int_{0}^{1} x (x^{2} + 1)^{2} d x

Expand

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

= 6 \cdot \int_{0}^{1} 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

= 6 (\int_{0}^{1} x^{5} d x + \int_{0}^{1} 2 x^{3} d x + \int_{0}^{1} x d x)

\int_{0}^{1} x^{5} d x = \frac{1}{6}

\int_{0}^{1} 2 x^{3} d x = \frac{1}{2}

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

= 6 (\frac{1}{6} + \frac{1}{2} + \frac{1}{2})

Simplify

6 (\frac{1}{6} + \frac{1}{2} + \frac{1}{2}) : 7

= 7

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\int_{- 4}^{0} 1

What is the integral from 0 to 1 of 6x(x^2+1)^2 ?
The integral from 0 to 1 of 6x(x^2+1)^2 is 7