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

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

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

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

ln ∣ 1 + arctan (x) ∣ + C

Solution steps

\int \frac{1}{( 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 = 1 + arctan (x)

= ln ∣ 1 + arctan (x) ∣

Add a constant to the solution

= ln ∣ 1 + arctan (x) ∣ + C

x \to 7 lim (e^{x} - e^{- x} - x)

9 x^{2} + 2 x, at x = 3

\int 2^{x} e^{x} d x

y = x^{3} - 3 x + 3, (2, 5)

\frac{d}{d x} (sin (0))

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