What is the integral of arcsec(4x) ?

The integral of arcsec(4x) is xarcsec(4x)-1/4 ln|4x+sqrt(16x^2-1)|+C

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integral of arcsec(4x)

\int arcsec (4 x) d x

x arcsec (4 x) - \frac{1}{4} ln 4 x + 16 x^{2} - 1 + C

Solution steps

\int arcsec (4 x) d x

Apply u-substitution

: \int arcsec (u) \frac{1}{4} d u

= \int arcsec (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \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}{4} (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}{4} (u arcsec (u) - ln u + u^{2} - 1)

Substitute back

u = 4 x

= \frac{1}{4} (4 x arcsec (4 x) - ln 4 x + (4 x)^{2} - 1)

Simplify

\frac{1}{4} (4 x arcsec (4 x) - ln 4 x + (4 x)^{2} - 1) : x arcsec (4 x) - \frac{1}{4} ln 4 x + 16 x^{2} - 1

= x arcsec (4 x) - \frac{1}{4} ln 4 x + 16 x^{2} - 1

Add a constant to the solution

= x arcsec (4 x) - \frac{1}{4} ln 4 x + 16 x^{2} - 1 + C

(e^{t} x + 1) d x

\frac{\partial}{\partial x} (2 y + x)

\int y - y d y

\frac{13}{ln ( x )}

y = \frac{1}{x}, at x = 4

What is the integral of arcsec(4x) ?
The integral of arcsec(4x) is xarcsec(4x)-1/4 ln|4x+sqrt(16x^2-1)|+C