What is the integral from 0 to 1 of 8x(x^2+1)^3 ?

The integral from 0 to 1 of 8x(x^2+1)^3 is 15

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integral from 0 to 1 of 8x(x^2+1)^3

\int_{0}^{1} 8 x (x^{2} + 1)^{3} d x

15

Solution steps

\int_{0}^{1} 8 x (x^{2} + 1)^{3} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{1} x (x^{2} + 1)^{3} d x

Expand

x (x^{2} + 1)^{3} : x^{7} + 3 x^{5} + 3 x^{3} + x

= 8 \cdot \int_{0}^{1} x^{7} + 3 x^{5} + 3 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

= 8 (\int_{0}^{1} x^{7} d x + \int_{0}^{1} 3 x^{5} d x + \int_{0}^{1} 3 x^{3} d x + \int_{0}^{1} x d x)

\int_{0}^{1} x^{7} d x = \frac{1}{8}

\int_{0}^{1} 3 x^{5} d x = \frac{1}{2}

\int_{0}^{1} 3 x^{3} d x = \frac{3}{4}

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

= 8 (\frac{1}{8} + \frac{1}{2} + \frac{3}{4} + \frac{1}{2})

Simplify

8 (\frac{1}{8} + \frac{1}{2} + \frac{3}{4} + \frac{1}{2}) : 15

= 15

\int_{2}^{7} x x + 2 d x

\int_{0}^{3} x^{2} + 1 d x

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

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

\int_{- 3}^{0} x + 3 + 9 - x^{2} d x

What is the integral from 0 to 1 of 8x(x^2+1)^3 ?
The integral from 0 to 1 of 8x(x^2+1)^3 is 15