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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Decimal Notation

2.29558 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{3} (u) d u

= \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} sec (u) d u

\int_{- \frac{π}{4}}^{\frac{π}{4}} sec (u) d u = ln (1 + 2) - ln (- 1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2)) : [\frac{1}{2} sec (u) tan (u)]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))

= [\frac{1}{2} sec (u) tan (u)]_{- \frac{π}{4}}^{\frac{π}{4}} + \frac{1}{2} (ln (1 + 2) - ln (- 1 + 2))

Compute the boundaries:

2

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

\int tan (x) sec (x) d x

y = arctan (ln (2) x^{4})

(- 3, 4), (4, 1)

\int_{- 2}^{1} \frac{1}{2} x^{2} + x + 1 d x

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

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