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

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

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

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

\frac{2 ( 1 + 2 )}{15}

Decimal

0.32189 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} \frac{( u - 1 ) u}{2} d u

= \int_{1}^{2} \frac{( u - 1 ) u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{2} (u - 1) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

= \frac{1}{2} \cdot \int_{1}^{2} u^{\frac{3}{2}} - u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{1}^{2} u^{\frac{3}{2}} d u - \int_{1}^{2} u d u)

\int_{1}^{2} u^{\frac{3}{2}} d u = \frac{8 2 - 2}{5}

\int_{1}^{2} u d u = \frac{4 2 - 2}{3}

= \frac{1}{2} (\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3})

Simplify

\frac{1}{2} (\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3}) : \frac{2 ( 1 + 2 )}{15}

= \frac{2 ( 1 + 2 )}{15}

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