What is the integral of (sec^2(y))/(a+btan(y)) ?

The integral of (sec^2(y))/(a+btan(y)) is 1/b ln|a+btan(y)|+C

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integral of (sec^2(y))/(a+btan(y))

\int \frac{sec ^{2} ( y )}{a + b tan ( y )} d y

\frac{1}{b} ln ∣ a + b tan (y) ∣ + C

Solution steps

\int \frac{sec ^{2} ( y )}{a + b tan ( y )} d y

Apply u-substitution

: \int \frac{1}{b u} d u

= \int \frac{1}{b u} d u

Take the constant out:

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

= \frac{1}{b} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{b} ln ∣ u ∣

Substitute back

u = a + b tan (y)

= \frac{1}{b} ln ∣ a + b tan (y) ∣

Add a constant to the solution

= \frac{1}{b} ln ∣ a + b tan (y) ∣ + C

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What is the integral of (sec^2(y))/(a+btan(y)) ?
The integral of (sec^2(y))/(a+btan(y)) is 1/b ln|a+btan(y)|+C