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))

\int tan^{2} (3 t) d t

\frac{\partial}{\partial x} (x \cdot a + b)

y = \frac{x - 1}{x - 2} + 1, (3, 3)

ln (x^{2} + y) - x y^{2} = 10

\frac{cos ( x )}{1 + a \cdot sin ( x )}

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)))