What is the integral of arcsec(5x) ?

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

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integral of arcsec(5x)

\int arcsec (5 x) d x

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

Solution steps

\int arcsec (5 x) d x

Apply u-substitution

: \int arcsec (u) \frac{1}{5} d u

= \int arcsec (u) \frac{1}{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 arcsec (u) d u

Apply Integration By Parts

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

= \frac{1}{5} (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}{5} (u arcsec (u) - ln u + u^{2} - 1)

Substitute back

u = 5 x

= \frac{1}{5} (5 x arcsec (5 x) - ln 5 x + (5 x)^{2} - 1)

Simplify

\frac{1}{5} (5 x arcsec (5 x) - ln 5 x + (5 x)^{2} - 1) : x arcsec (5 x) - \frac{1}{5} ln 5 x + 25 x^{2} - 1

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

Add a constant to the solution

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

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