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

The integral of 1/(2+tan(x)) is 1/5 ln|tan(x)+2|+1/5 (-1/2 ln(sec^2(x))+2arctan(tan(x)))+C

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

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

\frac{1}{5} ln ∣ tan (x) + 2 ∣ + \frac{1}{5} (- \frac{1}{2} ln (sec^{2} (x)) + 2 arctan (tan (x))) + C

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the partial fraction of

\frac{1}{( 2 + u ) ( 1 + u ^{2} )} : \frac{1}{5 ( u + 2 )} + \frac{- u + 2}{5 ( u ^{2} + 1 )}

= \int \frac{1}{5 ( u + 2 )} + \frac{- u + 2}{5 ( u ^{2} + 1 )} 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}{5 ( u + 2 )} d u + \int \frac{- u + 2}{5 ( u ^{2} + 1 )} d u

\int \frac{1}{5 ( u + 2 )} d u = \frac{1}{5} ln ∣ u + 2 ∣

\int \frac{- u + 2}{5 ( u ^{2} + 1 )} d u = \frac{1}{5} (- \frac{1}{2} ln u^{2} + 1 + 2 arctan (u))

= \frac{1}{5} ln ∣ u + 2 ∣ + \frac{1}{5} (- \frac{1}{2} ln u^{2} + 1 + 2 arctan (u))

Substitute back

u = tan (x)

= \frac{1}{5} ln ∣ tan (x) + 2 ∣ + \frac{1}{5} (- \frac{1}{2} ln tan^{2} (x) + 1 + 2 arctan (tan (x)))

\frac{1}{2} ln tan^{2} (x) + 1 = \frac{1}{2} ln (sec^{2} (x))

= \frac{1}{5} ln ∣ tan (x) + 2 ∣ + \frac{1}{5} (- \frac{1}{2} ln (sec^{2} (x)) + 2 arctan (tan (x)))

Add a constant to the solution

= \frac{1}{5} ln ∣ tan (x) + 2 ∣ + \frac{1}{5} (- \frac{1}{2} ln (sec^{2} (x)) + 2 arctan (tan (x))) + C

\int \frac{x + 4}{x ^{2} + 8 x - 20} d x

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

x \to 0 lim (10 - x^{2})

\int \frac{sec ^{2} ( x )}{3} d x

\int_{0}^{1} 2^{x} d x

What is the integral of 1/(2+tan(x)) ?
The integral of 1/(2+tan(x)) is 1/5 ln|tan(x)+2|+1/5 (-1/2 ln(sec^2(x))+2arctan(tan(x)))+C