What is the integral from 2 to 4 of 1/(sqrt(x^2-4)) ?

The integral from 2 to 4 of 1/(sqrt(x^2-4)) is ln(2+sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to 4 of 1/(sqrt(x^2-4))

\int_{2}^{4} \frac{1}{x ^{2} - 4} d x

ln (2 + 3)

Decimal Notation

1.31695 \dots

Solution steps

\int_{2}^{4} \frac{1}{x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{3}} sec (u) d u

= \int_{0}^{\frac{π}{3}} sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= [ln ∣ tan (u) + sec (u) ∣]_{0}^{\frac{π}{3}}

Compute the boundaries:

ln (2 + 3)

= ln (2 + 3)

\int_{1}^{3} 4 x d x

\int_{- 6}^{4} \frac{3}{x ^{4}} d x

\int_{0}^{8} x^{\frac{2}{3}} d x

\int_{- 1}^{0} \frac{1}{1 + e ^{x}} d x

\int_{0}^{\frac{3 π}{2}} sin (x) d x

What is the integral from 2 to 4 of 1/(sqrt(x^2-4)) ?
The integral from 2 to 4 of 1/(sqrt(x^2-4)) is ln(2+sqrt(3))