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

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

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integral of 1/(1+tan(u))

\int \frac{1}{1 + tan ( u )} d u

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

Solution steps

\int \frac{1}{1 + tan ( u )} d u

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

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

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

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

Substitute back

v = tan (u)

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

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

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

Add a constant to the solution

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

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