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

The integral from-x to x of sqrt(1+t^2) is xsqrt(1+x^2)+1/2 (ln|x+sqrt(1+x^2)|-ln|-x+sqrt(1+x^2)|)

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

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

x 1 + x^{2} + \frac{1}{2} (ln x + 1 + x^{2} - ln - x + 1 + x^{2})

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

\int_{a r c t a n (- x)}^{a r c t a n (x)} sec (u) d u = ln x + 1 + x^{2} - ln - x + 1 + x^{2}

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

Simplify

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

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

Compute the boundaries:

x 1 + x^{2}

= x 1 + x^{2} + \frac{1}{2} (ln x + 1 + x^{2} - ln - x + 1 + x^{2})

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