What is the integral from 0 to x of sqrt(1+t^2) ?

The integral from 0 to x of sqrt(1+t^2) is (xsqrt(x^2+1)+ln|x+sqrt(x^2+1)|)/2

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integral from 0 to x of sqrt(1+t^2)

\int_{0}^{x} 1 + t^{2} d t

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

Solution steps

\int_{0}^{x} 1 + t^{2} d t

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (x)} sec^{3} (u) d u

= \int_{0}^{a r c t a n (x)} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (x)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (x)} sec (u) d u

\int_{0}^{a r c t a n (x)} sec (u) d u = ln x + x^{2} + 1

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (x)} + \frac{1}{2} ln x + x^{2} + 1

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (x)} + \frac{1}{2} ln x + x^{2} + 1 : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (x)} + \frac{1}{2} ln x + x^{2} + 1

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (x)} + \frac{1}{2} ln x + x^{2} + 1

Compute the boundaries:

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

= \frac{1}{2} x x^{2} + 1 + \frac{1}{2} ln x + x^{2} + 1

Simplify

= \frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2}

\frac{d}{d x} (4 x^{\frac{1}{4}})

x \to 3 lim (x - 1)

\frac{\partial}{\partial x} (10 x^{2} y)

x \to 4 + lim (\frac{3}{( 4 - x ) ^{3}})

g (x) = 7 - 3 x

What is the integral from 0 to x of sqrt(1+t^2) ?
The integral from 0 to x of sqrt(1+t^2) is (xsqrt(x^2+1)+ln|x+sqrt(x^2+1)|)/2