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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

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

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