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

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

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integral of 1/(sqrt(y^2-9))

\int \frac{1}{y ^{2} - 9} d y

ln (\frac{1}{3} y^{2} - 9 + y) + C

Solution steps

\int \frac{1}{y ^{2} - 9} d y

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}{3} y)

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

Simplify

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

= ln (\frac{1}{3} y^{2} - 9 + y)

Add a constant to the solution

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

t^{3} sin (2 t)

x \cdot \frac{d y}{d x} - y = x^{2} sin (x)

\int y - 1 d y

\frac{3}{( s + 2 ) ^{2} + 9}

\int \frac{1}{9 sin ^{2} ( θ )} d θ

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