What is the integral of (sec^2(9sqrt(x)))/(sqrt(x)) ?

The integral of (sec^2(9sqrt(x)))/(sqrt(x)) is 2/9 tan(9sqrt(x))+C

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integral of (sec^2(9sqrt(x)))/(sqrt(x))

\int \frac{sec ^{2} ( 9 x )}{x} d x

\frac{2}{9} tan (9 x) + C

Solution steps

\int \frac{sec ^{2} ( 9 x )}{x} d x

Apply u-substitution

: \int \frac{2 sec ^{2} ( u )}{9} d u

= \int \frac{2 sec ^{2} ( u )}{9} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{2}{9} tan (u)

Substitute back

u = 9 x

= \frac{2}{9} tan (9 x)

Add a constant to the solution

= \frac{2}{9} tan (9 x) + C

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What is the integral of (sec^2(9sqrt(x)))/(sqrt(x)) ?
The integral of (sec^2(9sqrt(x)))/(sqrt(x)) is 2/9 tan(9sqrt(x))+C