What is the integral of+1/(sqrt(25x^2-36)) ?

The integral of+1/(sqrt(25x^2-36)) is 1/5 ln((|sqrt(25x^2-36)+5x|)/6)+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(25x^2-36))

\int + \frac{1}{25 x ^{2} - 36} d x

\frac{1}{5} ln (\frac{25 x ^{2} - 36 + 5 x}{6}) + C

Solution steps

\int \frac{1}{25 x ^{2} - 36} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5} d u

= \int \frac{sec ( u )}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{5}{6} x)

= \frac{1}{5} ln tan (arcsec (\frac{5}{6} x)) + sec (arcsec (\frac{5}{6} x))

Simplify

\frac{1}{5} ln tan (arcsec (\frac{5}{6} x)) + sec (arcsec (\frac{5}{6} x)) : \frac{1}{5} ln (\frac{25 x ^{2} - 36 + 5 x}{6})

= \frac{1}{5} ln (\frac{25 x ^{2} - 36 + 5 x}{6})

Add a constant to the solution

= \frac{1}{5} ln (\frac{25 x ^{2} - 36 + 5 x}{6}) + C

\int \frac{x + 5}{x ^{2} - 6 x - 15} d x

\int (2 x - 6) d x

\int \frac{x ^{2} - 4}{x + 1} d x

\int \frac{4 x - 16}{x + 2} d x

\int sec^{2} (3 x) tan^{4} (3 x) d x

What is the integral of+1/(sqrt(25x^2-36)) ?
The integral of+1/(sqrt(25x^2-36)) is 1/5 ln((|sqrt(25x^2-36)+5x|)/6)+C