What is the integral of 1/(sqrt(25x^2-4)) ?

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

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integral of 1/(sqrt(25x^2-4))

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

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

Solution steps

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

= \frac{1}{5} ln tan (arcsec (\frac{5}{2} x)) + sec (arcsec (\frac{5}{2} x))

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (e^{x} sin (y) + 3 y)

\int \frac{1}{x ^{2} + 3 x - 4} d x

\int \frac{x ^{2} + 1}{x ^{3}} d x

x \to 0 lim (4 - x)

\frac{\partial}{\partial x} (ln (x y) + ln (yz) + 5 ln (x z))

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