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

The integral from 0 to 1 of sqrt(4+x^2) is 4((sqrt(5))/8+1/2 ln((1+sqrt(5))/2))

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integral from 0 to 1 of sqrt(4+x^2)

\int_{0}^{1} 4 + x^{2} d x

4 (\frac{5}{8} + \frac{1}{2} ln (\frac{1 + 5}{2}))

Decimal

2.08045 \dots

Solution steps

\int_{0}^{1} 4 + x^{2} d x

Apply Trigonometric Substitution

= \int_{0}^{a r c t a n (\frac{1}{2})} 4 sec^{3} (u) d u

Take the constant out:

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

= 4 \cdot \int_{0}^{a r c t a n (\frac{1}{2})} sec^{3} (u) d u

Apply Integral Reduction

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{2})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{1}{2})} sec (u) d u

\int_{0}^{a r c t a n (\frac{1}{2})} sec (u) d u = ln (\frac{1}{2} + \frac{5}{2})

= 4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{2})} + \frac{1}{2} ln (\frac{1}{2} + \frac{5}{2})

Simplify

4 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{2})} + \frac{1}{2} ln (\frac{1}{2} + \frac{5}{2}) : 4 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{1}{2})} + \frac{1}{2} ln (\frac{1}{2} + \frac{5}{2}))

= 4 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{1}{2})} + \frac{1}{2} ln (\frac{1}{2} + \frac{5}{2}))

Compute the boundaries:

\frac{5}{8}

= 4 (\frac{5}{8} + \frac{1}{2} ln (\frac{1}{2} + \frac{5}{2}))

Simplify

= 4 (\frac{5}{8} + \frac{1}{2} ln (\frac{1 + 5}{2}))

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What is the integral from 0 to 1 of sqrt(4+x^2) ?
The integral from 0 to 1 of sqrt(4+x^2) is 4((sqrt(5))/8+1/2 ln((1+sqrt(5))/2))