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}^{1} 2 x^{2} - 2 x^{3} d x

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\int_{1}^{e} in x d x

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\int_{2}^{6} (x^{2} + 6) d 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