What is the integral from 0 to 7 of sqrt(1+x^2) ?

The integral from 0 to 7 of sqrt(1+x^2) is (35)/(sqrt(2))+1/2 ln(7+5sqrt(2))

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integral from 0 to 7 of sqrt(1+x^2)

\int_{0}^{7} 1 + x^{2} d x

\frac{35}{2} + \frac{1}{2} ln (7 + 52)

Decimal Notation

26.07079 \dots

Solution steps

\int_{0}^{7} 1 + x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (7)} sec^{3} (u) d u

= \int_{0}^{a r c t a n (7)} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (7)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (7)} sec (u) d u

\int_{0}^{a r c t a n (7)} sec (u) d u = ln (7 + 52)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (7)} + \frac{1}{2} ln (7 + 52)

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (7)} + \frac{1}{2} ln (7 + 52) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (7)} + \frac{1}{2} ln (7 + 52)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (7)} + \frac{1}{2} ln (7 + 52)

Compute the boundaries:

\frac{35}{2}

= \frac{35}{2} + \frac{1}{2} ln (7 + 52)

\frac{\partial}{\partial z} (cos (z))

\frac{d}{d x} (\frac{d y}{d x}) + y = cos (x)

\int sec (\frac{x}{3}) - csc (5 x) d x

\frac{d}{d x} (sec^{5} (x^{7} + 4 x)^{6})

x^{2020} + \frac{1}{3} x^{3} - 4 x^{2}

What is the integral from 0 to 7 of sqrt(1+x^2) ?
The integral from 0 to 7 of sqrt(1+x^2) is (35)/(sqrt(2))+1/2 ln(7+5sqrt(2))