What is the integral from-5 to 5 of sqrt(1+x^2) ?

The integral from-5 to 5 of sqrt(1+x^2) is 5sqrt(26)+1/2 (ln(5+sqrt(26))-ln(-5+sqrt(26)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from-5 to 5 of sqrt(1+x^2)

\int_{- 5}^{5} 1 + x^{2} d x

526 + \frac{1}{2} (ln (5 + 26) - ln (- 5 + 26))

Decimal

27.80753 \dots

Solution steps

\int_{- 5}^{5} 1 + x^{2} d x

Apply Trigonometric Substitution

= \int_{- a r c t a n (5)}^{a r c t a n (5)} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (5)}^{a r c t a n (5)} + \frac{1}{2} \cdot \int_{- a r c t a n (5)}^{a r c t a n (5)} sec (u) d u

\int_{- a r c t a n (5)}^{a r c t a n (5)} sec (u) d u = ln (5 + 26) - ln (- 5 + 26)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (5)}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (- 5 + 26))

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (5)}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (- 5 + 26)) : [\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (5)}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (- 5 + 26))

= [\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (5)}^{a r c t a n (5)} + \frac{1}{2} (ln (5 + 26) - ln (- 5 + 26))

Compute the boundaries:

526

= 526 + \frac{1}{2} (ln (5 + 26) - ln (- 5 + 26))

(e^{y})^{^{'}}

\int \frac{1 - x ^{2}}{2} d x

\frac{d}{d x} (lo g_{2} (x - 1) - 2)

d er i v a t i v e y = x^{2} e^{x} - 2 x e^{x} + 6 e^{x}

t an g e n t f (x) = 16 x, at x = 16

What is the integral from-5 to 5 of sqrt(1+x^2) ?
The integral from-5 to 5 of sqrt(1+x^2) is 5sqrt(26)+1/2 (ln(5+sqrt(26))-ln(-5+sqrt(26)))