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

The integral from 0 to 3 of sqrt(1+x^2) is (3sqrt(5))/(sqrt(2))+1/2 ln(3+sqrt(10))

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

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

\frac{3 5}{2} + \frac{1}{2} ln (3 + 10)

Decimal Notation

5.65263 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (3)} sec^{3} (u) d u

= \int_{0}^{a r c t a n (3)} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (3)} sec (u) d u

\int_{0}^{a r c t a n (3)} sec (u) d u = ln (3 + 10)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10)

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10)

Compute the boundaries:

\frac{3 5}{2}

= \frac{3 5}{2} + \frac{1}{2} ln (3 + 10)

x \to 6 + lim (6 x - 6)

y = \frac{x}{4} - \frac{4}{x}

\frac{( 2 )}{s ^{2} + 5 s + 3}

\frac{d}{d x} (a x + 4)

\frac{\partial}{\partial y} (\frac{y}{3})

What is the integral from 0 to 3 of sqrt(1+x^2) ?
The integral from 0 to 3 of sqrt(1+x^2) is (3sqrt(5))/(sqrt(2))+1/2 ln(3+sqrt(10))