What is the integral of 5/(sqrt(25x^2-9)) ?

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

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integral of 5/(sqrt(25x^2-9))

\int \frac{5}{25 x ^{2} - 9} d x

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

Solution steps

\int \frac{5}{25 x ^{2} - 9} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{25 x ^{2} - 9} d x

Apply Trigonometric Substitution

= 5 \cdot \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

= 5 \cdot \frac{1}{5} \cdot \int sec (u) d u

Use the common integral:

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

= 5 \cdot \frac{1}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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