What is the integral of 1/(sqrt(4x^2-81)) ?

The integral of 1/(sqrt(4x^2-81)) is 1/2 ln((|sqrt(4x^2-81)+2x|)/9)+C

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integral of 1/(sqrt(4x^2-81))

\int \frac{1}{4 x ^{2} - 81} d x

\frac{1}{2} ln (\frac{4 x ^{2} - 81 + 2 x}{9}) + C

Solution steps

\int \frac{1}{4 x ^{2} - 81} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{2} d u

= \int \frac{sec ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{2}{9} x)

= \frac{1}{2} ln tan (arcsec (\frac{2}{9} x)) + sec (arcsec (\frac{2}{9} x))

Simplify

\frac{1}{2} ln tan (arcsec (\frac{2}{9} x)) + sec (arcsec (\frac{2}{9} x)) : \frac{1}{2} ln (\frac{4 x ^{2} - 81 + 2 x}{9})

= \frac{1}{2} ln (\frac{4 x ^{2} - 81 + 2 x}{9})

Add a constant to the solution

= \frac{1}{2} ln (\frac{4 x ^{2} - 81 + 2 x}{9}) + C

\int \frac{x ^{3}}{25 + x ^{8}} d x

3 + v t - 4.9 t^{2}

\int_{0}^{5} \frac{1}{25 - x ^{2}} d x

\frac{\partial}{\partial y} (\frac{x ^{2}}{y ^{3}})

\frac{\partial}{\partial θ} (r (θ) cos^{2} (θ))

What is the integral of 1/(sqrt(4x^2-81)) ?
The integral of 1/(sqrt(4x^2-81)) is 1/2 ln((|sqrt(4x^2-81)+2x|)/9)+C