What is the integral of 1/(3+tan^2(x)) ?

The integral of 1/(3+tan^2(x)) is 1/2 arctan(tan(x))-1/(2sqrt(3))arctan((tan(x))/(sqrt(3)))+C

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integral of 1/(3+tan^2(x))

\int \frac{1}{3 + tan ^{2} ( x )} d x

\frac{1}{2} arctan (tan (x)) - \frac{1}{2 3} arctan (\frac{tan ( x )}{3}) + C

Solution steps

\int \frac{1}{3 + tan ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{( 1 + u ^{2} ) ( 3 + u ^{2} )} d u

= \int \frac{1}{( 1 + u ^{2} ) ( 3 + u ^{2} )} d u

Take the partial fraction of

\frac{1}{( 1 + u ^{2} ) ( 3 + u ^{2} )} : \frac{1}{2 ( u ^{2} + 1 )} - \frac{1}{2 ( u ^{2} + 3 )}

= \int \frac{1}{2 ( u ^{2} + 1 )} - \frac{1}{2 ( u ^{2} + 3 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2 ( u ^{2} + 1 )} d u - \int \frac{1}{2 ( u ^{2} + 3 )} d u

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

\int \frac{1}{2 ( u ^{2} + 3 )} d u = \frac{1}{2 3} arctan (\frac{u}{3})

= \frac{1}{2} arctan (u) - \frac{1}{2 3} arctan (\frac{u}{3})

Substitute back

u = tan (x)

= \frac{1}{2} arctan (tan (x)) - \frac{1}{2 3} arctan (\frac{tan ( x )}{3})

Add a constant to the solution

= \frac{1}{2} arctan (tan (x)) - \frac{1}{2 3} arctan (\frac{tan ( x )}{3}) + C

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What is the integral of 1/(3+tan^2(x)) ?
The integral of 1/(3+tan^2(x)) is 1/2 arctan(tan(x))-1/(2sqrt(3))arctan((tan(x))/(sqrt(3)))+C