What is the integral of 1/(sqrt(9x^2-25)) ?

The integral of 1/(sqrt(9x^2-25)) is 1/3 ln((|sqrt(9x^2-25)+3x|)/5)+C

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integral of 1/(sqrt(9x^2-25))

\int \frac{1}{9 x ^{2} - 25} d x

\frac{1}{3} ln (\frac{9 x ^{2} - 25 + 3 x}{5}) + C

Solution steps

\int \frac{1}{9 x ^{2} - 25} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3} d u

= \int \frac{sec ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{3}{5} x)

= \frac{1}{3} ln tan (arcsec (\frac{3}{5} x)) + sec (arcsec (\frac{3}{5} x))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{3}{5} x)) + sec (arcsec (\frac{3}{5} x)) : \frac{1}{3} ln (\frac{9 x ^{2} - 25 + 3 x}{5})

= \frac{1}{3} ln (\frac{9 x ^{2} - 25 + 3 x}{5})

Add a constant to the solution

= \frac{1}{3} ln (\frac{9 x ^{2} - 25 + 3 x}{5}) + C

\int \frac{1}{x + 1 + x - 1} d x

f (x) = x^{2} - 1, g (x) = - x + 2, 0, 1

\int (2 - x) 3^{(2 - x)^{2}} d x

\int_{0}^{3} \frac{1}{9 - x ^{2}} d x

\frac{d}{d x} (- A B^{- x} + C)

What is the integral of 1/(sqrt(9x^2-25)) ?
The integral of 1/(sqrt(9x^2-25)) is 1/3 ln((|sqrt(9x^2-25)+3x|)/5)+C