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

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

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

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

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

Solution steps

\int \frac{1}{16 x ^{2} - 25} 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}{5} x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (π)

2 t \frac{d y}{d t} - 2 y = t

\frac{\partial}{\partial x} (ln (x + 1))

\int_{- 7}^{7} (e^{x} - e^{- x})^{2} d x

\frac{d}{d x} (\frac{x + x}{x ^{2}})

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