What is the integral of 1/(sqrt(16x^2-81)) ?

The integral of 1/(sqrt(16x^2-81)) is 1/4 ln((|sqrt(16x^2-81)+4x|)/9)+C

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integral of 1/(sqrt(16x^2-81))

\int \frac{1}{16 x ^{2} - 81} d x

\frac{1}{4} ln (\frac{16 x ^{2} - 81 + 4 x}{9}) + C

Solution steps

\int \frac{1}{16 x ^{2} - 81} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4} d u

= \int \frac{sec ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{4} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{4}{9} x)

= \frac{1}{4} ln tan (arcsec (\frac{4}{9} x)) + sec (arcsec (\frac{4}{9} x))

Simplify

\frac{1}{4} ln tan (arcsec (\frac{4}{9} x)) + sec (arcsec (\frac{4}{9} x)) : \frac{1}{4} ln (\frac{16 x ^{2} - 81 + 4 x}{9})

= \frac{1}{4} ln (\frac{16 x ^{2} - 81 + 4 x}{9})

Add a constant to the solution

= \frac{1}{4} ln (\frac{16 x ^{2} - 81 + 4 x}{9}) + C

\int ln (\frac{3}{x}) d x

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

f (x) = ln (ln (ln (x)))

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

n = 0 \sum \infty \frac{n !}{10 0 ^{n}}

What is the integral of 1/(sqrt(16x^2-81)) ?
The integral of 1/(sqrt(16x^2-81)) is 1/4 ln((|sqrt(16x^2-81)+4x|)/9)+C