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

The integral from 0 to 4 of sqrt(1+x^2) is 2sqrt(17)+1/2 ln(4+sqrt(17))

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

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

217 + \frac{1}{2} ln (4 + 17)

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (4)} sec^{3} (u) d u

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

Apply Integral Reduction

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

\int_{0}^{a r c t a n (4)} sec (u) d u = ln (4 + 17)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17)

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17)

Compute the boundaries:

217

= 217 + \frac{1}{2} ln (4 + 17)

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What is the integral from 0 to 4 of sqrt(1+x^2) ?
The integral from 0 to 4 of sqrt(1+x^2) is 2sqrt(17)+1/2 ln(4+sqrt(17))