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

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

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

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

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

Decimal

52.14159 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

352

= 352 + \frac{1}{2} (ln (7 + 52) - ln (- 7 + 52))

x \to 0 lim (\frac{4}{3 + e ^{\frac{1}{x}}})

\frac{d}{d x} (sin (2 x + 3 y))

\frac{\partial}{\partial y} (ln (x^{2} + y))

x^{4} - 2 x^{2}, 2 x^{2}, [0, 2]

f (x) = \frac{8 x}{x + 7}, (2, 1.7777700000000 \dots)

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