What is the integral from 0 to 1 of x^7sqrt(x^4+1) ?

The integral from 0 to 1 of x^7sqrt(x^4+1) is (sqrt(2)+1)/(15)

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integral from 0 to 1 of x^7sqrt(x^4+1)

\int_{0}^{1} x^{7} x^{4} + 1 d x

\frac{2 + 1}{15}

Decimal

0.16094 \dots

Solution steps

\int_{0}^{1} x^{7} x^{4} + 1 d x

Apply u-substitution

: \int_{0}^{1} \frac{u ^{\frac{1}{7}} u ^{\frac{4}{7}} + 1}{7} d u

= \int_{0}^{1} \frac{u ^{\frac{1}{7}} u ^{\frac{4}{7}} + 1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int_{0}^{1} u^{\frac{1}{7}} u^{\frac{4}{7}} + 1 d u

Apply u-substitution

: \int_{1}^{2} \frac{7 ( v - 1 ) v}{4} d v

= \frac{1}{7} \cdot \int_{1}^{2} \frac{7 ( v - 1 ) v}{4} d v

Take the constant out:

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

= \frac{1}{7} \cdot \frac{7}{4} \cdot \int_{1}^{2} (v - 1) v d v

Expand

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

= \frac{1}{7} \cdot \frac{7}{4} \cdot \int_{1}^{2} v^{\frac{3}{2}} - v d v

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}{7} \cdot \frac{7}{4} (\int_{1}^{2} v^{\frac{3}{2}} d v - \int_{1}^{2} v d v)

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

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

= \frac{1}{7} \cdot \frac{7}{4} (\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3})

Simplify

\frac{1}{7} \cdot \frac{7}{4} (\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3}) : \frac{2 + 1}{15}

= \frac{2 + 1}{15}

\frac{d y}{d x} = \frac{e ^{x} - 1}{e ^{y}}

\int_{1}^{4} \frac{x ^{2} - 1}{x} d x

x \to 9 lim (\frac{5 x}{x - 9})

f (x) = ((x^{2} + 7) (5 - x))

\frac{\partial}{\partial x} (e^{x - y})

What is the integral from 0 to 1 of x^7sqrt(x^4+1) ?
The integral from 0 to 1 of x^7sqrt(x^4+1) is (sqrt(2)+1)/(15)