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

The integral from 0 to 1 of 2x(x^2+1)^3 is 15/4

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

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

\frac{15}{4}

Decimal Notation

3.75

Solution steps

\int_{0}^{1} 2 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

= 2 \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

= 2 \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

= 2 (\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}

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

Simplify

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

= \frac{15}{4}

\int_{- 2}^{2} (x + 3)^{2} d x

\int_{1}^{e} \frac{ln ( x ^{2} )}{x ^{2}} d x

\int_{2}^{0} - 4 x^{4} + 8 x^{3} d x

\int_{- 1}^{1} ∣ e^{x} - 1 ∣ d x

\int_{2}^{4} 3 x^{2} d x

What is the integral from 0 to 1 of 2x(x^2+1)^3 ?
The integral from 0 to 1 of 2x(x^2+1)^3 is 15/4