What is the integral of 1/(sqrt(x^2-225)) ?

The integral of 1/(sqrt(x^2-225)) is ln(1/15 |sqrt(x^2-225)+x|)+C

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integral of 1/(sqrt(x^2-225))

\int \frac{1}{x ^{2} - 225} d x

ln (\frac{1}{15} x^{2} - 225 + x) + C

Solution steps

\int \frac{1}{x ^{2} - 225} 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}{15} x)

= ln tan (arcsec (\frac{1}{15} x)) + sec (arcsec (\frac{1}{15} x))

Simplify

ln tan (arcsec (\frac{1}{15} x)) + sec (arcsec (\frac{1}{15} x)) : ln (\frac{1}{15} x^{2} - 225 + x)

= ln (\frac{1}{15} x^{2} - 225 + x)

Add a constant to the solution

= ln (\frac{1}{15} x^{2} - 225 + x) + C

\frac{\partial}{\partial s} (s + 2 t - u)

\frac{d}{d x} ((x - 1) (x - 2) (x - 3))

\frac{d P}{d t} = P (3 - P) - \frac{9}{4}

\int_{3}^{8} t^{3} ln (2 t) d t

\frac{1}{9 s - 1}

What is the integral of 1/(sqrt(x^2-225)) ?
The integral of 1/(sqrt(x^2-225)) is ln(1/15 |sqrt(x^2-225)+x|)+C