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)))

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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 Notation

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

= \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))

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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)))