What is the integral of arcsec(3x) ?

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

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integral of arcsec(3x)

\int arcsec (3 x) d x

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

Solution steps

\int arcsec (3 x) d x

Apply u-substitution

= \int arcsec (u) \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 arcsec (u) d u

Apply Integration By Parts

= \frac{1}{3} (u arcsec (u) - \int \frac{u}{u ^{2} u ^{2} - 1} d u)

\int \frac{u}{u ^{2} u ^{2} - 1} d u = ln u + u^{2} - 1

= \frac{1}{3} (u arcsec (u) - ln u + u^{2} - 1)

Substitute back

u = 3 x

= \frac{1}{3} (3 x arcsec (3 x) - ln 3 x + (3 x)^{2} - 1)

Simplify

\frac{1}{3} (3 x arcsec (3 x) - ln 3 x + (3 x)^{2} - 1) : x arcsec (3 x) - \frac{1}{3} ln 3 x + 9 x^{2} - 1

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

Add a constant to the solution

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

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