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

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

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

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

2 arctan (x) + 2 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u + 2} d u

= \int \frac{1}{u + 2} d u

Apply u-substitution

: \int 2 d v

= \int 2 d v

Integral of a constant:

\int a d x = a x

= 2 v

Substitute back

= 2 arctan (x) + 2

Add a constant to the solution

= 2 arctan (x) + 2 + C

\int \frac{2 v}{1 + v ^{2}} d v

\int sec (3 x + 5) tan (3 x + 5) d x

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

\int \frac{a + x}{a ^{2} - x ^{2}} d x

\int x (1 - x)^{5} d x

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