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

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

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

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

ln ∣ arctan (x) ∣ + C

Solution steps

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

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 (x)

= ln ∣ arctan (x) ∣

Add a constant to the solution

= ln ∣ arctan (x) ∣ + C

\int x^{2} \cdot cos (2 x) d x

\frac{d}{d x} (- \frac{2}{cos ( 2 x )} + \frac{8}{3})

\frac{\partial}{\partial x} (9 x^{2} cos (y^{6}))

\frac{d}{d x} (4 3 x + 2 ln (x) - 3 e^{x})

\frac{d}{d x} (\frac{5 x ^{4} + 2 x}{2 x ^{9}})

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