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

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

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

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

\frac{7}{3}

Decimal Notation

2.33333 \dots

Solution steps

\int_{0}^{1} 2 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

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

Expand

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

= 2 \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

= 2 (\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}

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

Simplify

2 (\frac{1}{6} + \frac{1}{2} + \frac{1}{2}) : \frac{7}{3}

= \frac{7}{3}

f (x) = \frac{6}{x ^{2}}

x, 64

\int + \frac{x}{3 - x ^{4}} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 1

\frac{\partial}{\partial x} (x^{2} + x y^{2} + y^{2})

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