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

The integral from 0 to 2 of sqrt(1+x^2) is sqrt(5)+1/2 ln(2+sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

5 + \frac{1}{2} ln (2 + 5)

Decimal

2.95788 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (2)} sec^{3} (u) d u

= \int_{0}^{a r c t a n (2)} sec^{3} (u) d u

Apply Integral Reduction

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

\int_{0}^{a r c t a n (2)} sec (u) d u = ln (2 + 5)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)

Compute the boundaries:

5

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

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

\int \frac{( x + 2 ) ^{2}}{3 3 - x} d x

\frac{s}{( s + 4 ) ^{2}}

n = 1 \sum \infty \frac{9 n}{7 n + 1}

y = x, y = 0, y = x - 2

What is the integral from 0 to 2 of sqrt(1+x^2) ?
The integral from 0 to 2 of sqrt(1+x^2) is sqrt(5)+1/2 ln(2+sqrt(5))