What is the integral of 1/(sqrt(x^2-441)) ?

The integral of 1/(sqrt(x^2-441)) is ln(1/21 |sqrt(x^2-441)+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^2-441))

\int \frac{1}{x ^{2} - 441} d x

ln (\frac{1}{21} x^{2} - 441 + x) + C

Solution steps

\int \frac{1}{x ^{2} - 441} 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}{21} x)

= ln tan (arcsec (\frac{1}{21} x)) + sec (arcsec (\frac{1}{21} x))

Simplify

ln tan (arcsec (\frac{1}{21} x)) + sec (arcsec (\frac{1}{21} x)) : ln (\frac{1}{21} x^{2} - 441 + x)

= ln (\frac{1}{21} x^{2} - 441 + x)

Add a constant to the solution

= ln (\frac{1}{21} x^{2} - 441 + x) + C

\int - 16 sin (4 x - \frac{π}{2}) d x

n = 1 \sum \infty \frac{1}{n} - \frac{1}{n + 2}

\int \frac{21 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

y^{^{'}} = 3 y, y (2) = 4

\frac{d}{d x} (1 + \frac{1}{x} + \frac{1}{x ^{2}})

What is the integral of 1/(sqrt(x^2-441)) ?
The integral of 1/(sqrt(x^2-441)) is ln(1/21 |sqrt(x^2-441)+x|)+C