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

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

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

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

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

Decimal Notation

1.14779 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

= \int_{0}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2) : [\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

Compute the boundaries:

\frac{1}{2}

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

Simplify

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

(x - 3) (x^{2} + 6)

4 x^{3} y + x^{4} y^{^{'}} = - 6 sec^{2} (x) tan (x)

\frac{\partial}{\partial x} (A e^{\frac{1}{x}})

\int_{0}^{6} 2 π (y + 2) (3 - \frac{y}{2}) d y

\frac{d y}{d x} - \frac{y}{x} = 3 x e^{x}

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