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

The integral from 2 to x of sqrt(1+t^2) is -sqrt(5)+(xsqrt(1+x^2)+ln|x+sqrt(1+x^2)|-ln(2+sqrt(5)))/2

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

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

- 5 + \frac{x 1 + x ^{2} + ln x + 1 + x ^{2} - ln ( 2 + 5 )}{2}

Solution steps

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

Apply Trigonometric Substitution

: \int_{a r c t a n (2)}^{a r c t a n (x)} sec^{3} (u) d u

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

\int_{a r c t a n (2)}^{a r c t a n (x)} sec (u) d u = ln x + 1 + x^{2} - ln (2 + 5)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (2 + 5))

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (2 + 5)) : [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (2 + 5))

= [\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (x)} + \frac{1}{2} (ln x + 1 + x^{2} - ln (2 + 5))

Compute the boundaries:

\frac{1}{2} x 1 + x^{2} - 5

= \frac{1}{2} x 1 + x^{2} - 5 + \frac{1}{2} (ln x + 1 + x^{2} - ln (2 + 5))

Simplify

= - 5 + \frac{x 1 + x ^{2} + ln x + 1 + x ^{2} - ln ( 2 + 5 )}{2}

y = (a x + b)^{6}

\int \frac{1}{x ^{2} + 16} d x

z \to - i lim (\frac{im ( z )}{z})

\frac{d}{d x} (\frac{2}{3} e^{x})

100 q + 100

What is the integral from 2 to x of sqrt(1+t^2) ?
The integral from 2 to x of sqrt(1+t^2) is -sqrt(5)+(xsqrt(1+x^2)+ln|x+sqrt(1+x^2)|-ln(2+sqrt(5)))/2