What is the integral of 1/(sqrt(9x^2-64)) ?

The integral of 1/(sqrt(9x^2-64)) is 1/3 ln((|sqrt(9x^2-64)+3x|)/8)+C

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integral of 1/(sqrt(9x^2-64))

\int \frac{1}{9 x ^{2} - 64} d x

\frac{1}{3} ln (\frac{9 x ^{2} - 64 + 3 x}{8}) + C

Solution steps

\int \frac{1}{9 x ^{2} - 64} 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}{8} x)

= \frac{1}{3} ln tan (arcsec (\frac{3}{8} x)) + sec (arcsec (\frac{3}{8} x))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{3}{8} x)) + sec (arcsec (\frac{3}{8} x)) : \frac{1}{3} ln (\frac{9 x ^{2} - 64 + 3 x}{8})

= \frac{1}{3} ln (\frac{9 x ^{2} - 64 + 3 x}{8})

Add a constant to the solution

= \frac{1}{3} ln (\frac{9 x ^{2} - 64 + 3 x}{8}) + C

\int 0.4 x^{2} d x

y = x (x + 1)

x \to - 3 lim (\frac{( x + 4 )}{x + 3})

\int \frac{r ^{2}}{r + 4} d r

\frac{x - 1}{x + 2}

What is the integral of 1/(sqrt(9x^2-64)) ?
The integral of 1/(sqrt(9x^2-64)) is 1/3 ln((|sqrt(9x^2-64)+3x|)/8)+C