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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

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{\partial}{\partial y} (x^{2} y + y^{2} z + x z^{2})

f (x) = \frac{6 x - 1}{x ^{3}}

\int_{3}^{5} x^{3} d x

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

x, 2 x 81 - x^{2}

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