What is the integral of 1/(sqrt(x^2+100)) ?

The integral of 1/(sqrt(x^2+100)) is ln(1/10 |x+sqrt(100+x^2)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(x^2+100))

\int \frac{1}{x ^{2} + 100} d x

ln (\frac{1}{10} x + 100 + x^{2}) + C

Solution steps

\int \frac{1}{x ^{2} + 100} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

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

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{10} x)

= ln tan (arctan (\frac{1}{10} x)) + sec (arctan (\frac{1}{10} x))

Simplify

ln tan (arctan (\frac{1}{10} x)) + sec (arctan (\frac{1}{10} x)) : ln (\frac{1}{10} x + 100 + x^{2})

= ln (\frac{1}{10} x + 100 + x^{2})

Add a constant to the solution

= ln (\frac{1}{10} x + 100 + x^{2}) + C

t y^{^{'}} = 2 y + 2 t^{3} sin (3 t), y (\frac{π}{3}) = 0

x \to 0 - lim (\frac{x + 3}{x})

\frac{d}{d x} (\frac{x ^{4} + 2 x ^{2}}{x ^{2}})

3 t^{2} - 12

What is the integral of 1/(sqrt(x^2+100)) ?
The integral of 1/(sqrt(x^2+100)) is ln(1/10 |x+sqrt(100+x^2)|)+C