What is the integral of 1/(sqrt(25x^2-16)) ?

The integral of 1/(sqrt(25x^2-16)) is 1/5 ln((|sqrt(25x^2-16)+5x|)/4)+C

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integral of 1/(sqrt(25x^2-16))

\int \frac{1}{25 x ^{2} - 16} d x

\frac{1}{5} ln (\frac{25 x ^{2} - 16 + 5 x}{4}) + C

Solution steps

\int \frac{1}{25 x ^{2} - 16} 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}{4} x)

= \frac{1}{5} ln tan (arcsec (\frac{5}{4} x)) + sec (arcsec (\frac{5}{4} x))

Simplify

\frac{1}{5} ln tan (arcsec (\frac{5}{4} x)) + sec (arcsec (\frac{5}{4} x)) : \frac{1}{5} ln (\frac{25 x ^{2} - 16 + 5 x}{4})

= \frac{1}{5} ln (\frac{25 x ^{2} - 16 + 5 x}{4})

Add a constant to the solution

= \frac{1}{5} ln (\frac{25 x ^{2} - 16 + 5 x}{4}) + C

x \to 0 + lim (x ln (3 x))

\frac{\partial}{\partial z} (- 2 x y)

\int 3 x^{2} cos (6 x) d x

\frac{d}{d x} (ln (x^{2}) - \frac{1}{x ^{2}} - 5)

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 5})

What is the integral of 1/(sqrt(25x^2-16)) ?
The integral of 1/(sqrt(25x^2-16)) is 1/5 ln((|sqrt(25x^2-16)+5x|)/4)+C