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

The integral of 1/(sqrt(x^2-100)) is ln(1/10 |sqrt(x^2-100)+x|)+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^{2} - 100 + x) + 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 = arcsec (\frac{1}{10} x)

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x}{1 - 16 x ^{2}} d x

\frac{d}{d x} (2 sin (x) - sin (2 x))

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

\int 2 cot^{2} (x) csc (x) d x

sin (x), x = 0

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