What is the integral of (sec^4(ln(x)))/x ?

The integral of (sec^4(ln(x)))/x is tan(ln(x))+(tan^3(ln(x)))/3+C

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integral of (sec^4(ln(x)))/x

\int \frac{sec ^{4} ( ln ( x ) )}{x} d x

tan (ln (x)) + \frac{tan ^{3} ( ln ( x ) )}{3} + C

Solution steps

\int \frac{sec ^{4} ( ln ( x ) )}{x} d x

Apply u-substitution

= \int sec^{4} (u) d u

Rewrite using trig identities

= \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

= \int 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d v + \int v^{2} d v

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= v + \frac{v ^{3}}{3}

Substitute back

= tan (ln (x)) + \frac{tan ^{3} ( ln ( x ) )}{3}

Add a constant to the solution

= tan (ln (x)) + \frac{tan ^{3} ( ln ( x ) )}{3} + C

What is the integral of (sec^4(ln(x)))/x ?
The integral of (sec^4(ln(x)))/x is tan(ln(x))+(tan^3(ln(x)))/3+C