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}

\int_{0}^{5} π (2 x)^{2} d x

\int \frac{( sec ^{2} ( x ) )}{tan ^{2} ( x )} d x

\frac{d}{d x} y = (\frac{1}{cos ^{2} ( x )})

x \to - 9 lim (\frac{8 x + 72}{∣ x + 9 ∣})

\frac{( 2 \cdot s )}{s ^{2} + 6 s + 13}

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