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

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

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

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

ln ∣ arctan (y) ∣ + C

Solution steps

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

y = \frac{1}{14 x ^{2}}, y = x, y = \frac{x}{12}

n = 1 \sum \infty n^{2} + 1

2 x y - y^{2} = - 1

\frac{d}{d x} (cos^{4} (4 x^{2} + 1))

\int (3 x^{2} + 4 x - 6) d x

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