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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int_{\frac{π}{4}}^{a r c t a n (- x)} sec^{3} (u) d u

= \int_{\frac{π}{4}}^{a r c t a n (- x)} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (- x)} + \frac{1}{2} \cdot \int_{\frac{π}{4}}^{a r c t a n (- x)} sec (u) d u

\int_{\frac{π}{4}}^{a r c t a n (- x)} sec (u) d u = ln - x + 1 + x^{2} - ln (1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (- x)} + \frac{1}{2} (ln - x + 1 + x^{2} - ln (1 + 2))

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (- x)} + \frac{1}{2} (ln - x + 1 + x^{2} - ln (1 + 2)) : [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (- x)} + \frac{1}{2} (ln - x + 1 + x^{2} - ln (1 + 2))

= [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (- x)} + \frac{1}{2} (ln - x + 1 + x^{2} - ln (1 + 2))

Compute the boundaries:

- \frac{1}{2} x 1 + x^{2} - \frac{1}{2}

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

Simplify

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

\int_{- 3}^{3} 18 - x^{2} d x

\int_{0}^{1} e^{- s t} (t - 1) d t

\int_{- 2}^{0} \frac{1}{( x + 1 ) ^{\frac{1}{3}}} d x

\int_{0}^{3} 2 x d x

\int_{0}^{64} 3 x d x

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