What is the integral of 1/(sqrt(25x^2-36)) ?

The integral of 1/(sqrt(25x^2-36)) is 1/5 ln((|sqrt(25x^2-36)+5x|)/6)+C

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integral of 1/(sqrt(25x^2-36))

\int \frac{1}{25 x ^{2} - 36} d x

\frac{1}{5} ln (\frac{25 x ^{2} - 36 + 5 x}{6}) + C

Solution steps

\int \frac{1}{25 x ^{2} - 36} 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}{6} x)

= \frac{1}{5} ln tan (arcsec (\frac{5}{6} x)) + sec (arcsec (\frac{5}{6} x))

Simplify

\frac{1}{5} ln tan (arcsec (\frac{5}{6} x)) + sec (arcsec (\frac{5}{6} x)) : \frac{1}{5} ln (\frac{25 x ^{2} - 36 + 5 x}{6})

= \frac{1}{5} ln (\frac{25 x ^{2} - 36 + 5 x}{6})

Add a constant to the solution

= \frac{1}{5} ln (\frac{25 x ^{2} - 36 + 5 x}{6}) + C

3 ln (3 x)

\int \frac{3 x ^{2} - 21 x + 32}{x ^{3} - 8 x ^{2} + 16 x} d x

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

\frac{\partial}{\partial w} ((z + w) x^{\frac{1}{2}})

f (x) = sin (\frac{1}{x})

What is the integral of 1/(sqrt(25x^2-36)) ?
The integral of 1/(sqrt(25x^2-36)) is 1/5 ln((|sqrt(25x^2-36)+5x|)/6)+C