What is the integral of 1/((arctan(y))(1+y^2)) ?

The integral of 1/((arctan(y))(1+y^2)) is ln|arctan(y)|+C

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integral of 1/((arctan(y))(1+y^2))

\int \frac{1}{( arctan ( y ) ) ( 1 + y ^{2} )} d y

ln ∣ arctan (y) ∣ + C

Solution steps

\int \frac{1}{arctan ( y ) ( 1 + y ^{2} )} d y

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = arctan (y)

= ln ∣ arctan (y) ∣

Add a constant to the solution

= ln ∣ arctan (y) ∣ + C

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What is the integral of 1/((arctan(y))(1+y^2)) ?
The integral of 1/((arctan(y))(1+y^2)) is ln|arctan(y)|+C