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

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

Solution steps

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

Apply Trigonometric Substitution

= \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 = arctan (\frac{5}{6} x)

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

Simplify

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

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

Add a constant to the solution

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

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