What is the integral of 1/(sqrt(9z^2-25)) ?

The integral of 1/(sqrt(9z^2-25)) is 1/3 ln((|sqrt(9z^2-25)+3z|)/5)+C

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integral of 1/(sqrt(9z^2-25))

\int \frac{1}{9 z ^{2} - 25} d z

\frac{1}{3} ln (\frac{9 z ^{2} - 25 + 3 z}{5}) + C

Solution steps

\int \frac{1}{9 z ^{2} - 25} d z

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3} d u

= \int \frac{sec ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{3}{5} z)

= \frac{1}{3} ln tan (arcsec (\frac{3}{5} z)) + sec (arcsec (\frac{3}{5} z))

Simplify

\frac{1}{3} ln tan (arcsec (\frac{3}{5} z)) + sec (arcsec (\frac{3}{5} z)) : \frac{1}{3} ln (\frac{9 z ^{2} - 25 + 3 z}{5})

= \frac{1}{3} ln (\frac{9 z ^{2} - 25 + 3 z}{5})

Add a constant to the solution

= \frac{1}{3} ln (\frac{9 z ^{2} - 25 + 3 z}{5}) + C

y = 2 t (3 t^{2} - 5)^{4}

\int e^{5 x} cos (9 x) d x

f (x) = \frac{2}{x}, at x = 9

\frac{\partial}{\partial z} (- \frac{1}{( y + 2 z ) ^{2}})

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

What is the integral of 1/(sqrt(9z^2-25)) ?
The integral of 1/(sqrt(9z^2-25)) is 1/3 ln((|sqrt(9z^2-25)+3z|)/5)+C