What is the integral of 1/(sqrt(25x^2-64)) ?

The integral of 1/(sqrt(25x^2-64)) is 1/5 ln((|sqrt(25x^2-64)+5x|)/8)+C

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integral of 1/(sqrt(25x^2-64))

\int \frac{1}{25 x ^{2} - 64} d x

\frac{1}{5} ln (\frac{25 x ^{2} - 64 + 5 x}{8}) + C

Solution steps

\int \frac{1}{25 x ^{2} - 64} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5} d u

= \int \frac{sec ( u )}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{5}{8} x)

= \frac{1}{5} ln tan (arcsec (\frac{5}{8} x)) + sec (arcsec (\frac{5}{8} x))

Simplify

\frac{1}{5} ln tan (arcsec (\frac{5}{8} x)) + sec (arcsec (\frac{5}{8} x)) : \frac{1}{5} ln (\frac{25 x ^{2} - 64 + 5 x}{8})

= \frac{1}{5} ln (\frac{25 x ^{2} - 64 + 5 x}{8})

Add a constant to the solution

= \frac{1}{5} ln (\frac{25 x ^{2} - 64 + 5 x}{8}) + C

f (x) = x^{7}

2 t sin (t)

\frac{d}{d x} (\frac{3 x ^{2} + 2 y ^{2}}{( x - 2 y ) ^{2}})

f (x) = x, at x = \frac{1}{9}

\frac{d}{d x} (sin (2 x))

What is the integral of 1/(sqrt(25x^2-64)) ?
The integral of 1/(sqrt(25x^2-64)) is 1/5 ln((|sqrt(25x^2-64)+5x|)/8)+C