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_{0}^{8} (3 x^{2} + x + 6) d x

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\int_{1}^{x^{3}} \frac{g ( t )}{4 + t ^{2}} d t

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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