What is the integral of 1/(sqrt(49x^2-81)) ?

The integral of 1/(sqrt(49x^2-81)) is 1/7 ln((|sqrt(49x^2-81)+7x|)/9)+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(49x^2-81))

\int \frac{1}{49 x ^{2} - 81} d x

\frac{1}{7} ln (\frac{49 x ^{2} - 81 + 7 x}{9}) + C

Solution steps

\int \frac{1}{49 x ^{2} - 81} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{7} d u

= \int \frac{sec ( u )}{7} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{7} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{7} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{7}{9} x)

= \frac{1}{7} ln tan (arcsec (\frac{7}{9} x)) + sec (arcsec (\frac{7}{9} x))

Simplify

\frac{1}{7} ln tan (arcsec (\frac{7}{9} x)) + sec (arcsec (\frac{7}{9} x)) : \frac{1}{7} ln (\frac{49 x ^{2} - 81 + 7 x}{9})

= \frac{1}{7} ln (\frac{49 x ^{2} - 81 + 7 x}{9})

Add a constant to the solution

= \frac{1}{7} ln (\frac{49 x ^{2} - 81 + 7 x}{9}) + C

x \to (0) + lim (3 x^{(\frac{x}{2})})

\frac{d}{d x} ((5 x^{2} + 3) (4 x - 3))

\frac{d}{d x} (14 x e^{x})

\frac{\partial}{\partial x} (4 ln (x + z))

\int (x + 5) (x - 3) d x

What is the integral of 1/(sqrt(49x^2-81)) ?
The integral of 1/(sqrt(49x^2-81)) is 1/7 ln((|sqrt(49x^2-81)+7x|)/9)+C