What is the integral of 1/(ssqrt((4s)^2-9)) ?

The integral of 1/(ssqrt((4s)^2-9)) is 1/3 arcsec(4/3 s)+C

integral of 1/(ssqrt((4s)^2-9))

\int \frac{1}{s ( 4 s ) ^{2} - 9} d s

\frac{1}{3} arcsec (\frac{4}{3} s) + C

Solution steps

\int \frac{1}{s ( 4 s ) ^{2} - 9} d s

(4 s)^{2} = 16 s^{2}

= \int \frac{1}{s 16 s ^{2} - 9} d s

Apply Trigonometric Substitution

= \int \frac{1}{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 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{4}{3} s)

= \frac{1}{3} \cdot 1 \cdot arcsec (\frac{4}{3} s)

Simplify

= \frac{1}{3} arcsec (\frac{4}{3} s)

Add a constant to the solution

= \frac{1}{3} arcsec (\frac{4}{3} s) + C

\frac{\partial}{\partial x} (\frac{3 r + s}{t})

t an g e n t y = sin (x), at x = \frac{π}{4}

\int \frac{1}{( x + 1 ) ^{2} + 4} d x

\int x^{2} x^{3} + 33 d x

\frac{d y}{d x} + y^{5} x + 2 y = 0

What is the integral of 1/(ssqrt((4s)^2-9)) ?
The integral of 1/(ssqrt((4s)^2-9)) is 1/3 arcsec(4/3 s)+C