What is the integral of arctan(sqrt(u+1)) ?

The integral of arctan(sqrt(u+1)) is uarctan(sqrt(u+1))+2arctan(sqrt(u+1))-sqrt(u+1)+C

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integral of arctan(sqrt(u+1))

\int arctan (u + 1) d u

u arctan (u + 1) + 2 arctan (u + 1) - u + 1 + C

Solution steps

\int arctan (u + 1) d u

Apply u-substitution

: \int arctan (v) d v

= \int arctan (v) d v

Apply u-substitution

: \int arctan (w) \cdot 2 w d w

= \int arctan (w) \cdot 2 w d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int arctan (w) w d w

Apply Integration By Parts

: \frac{1}{2} w^{2} arctan (w) - \int \frac{w ^{2}}{2 ( w ^{2} + 1 )} d w

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

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

= 2 (\frac{1}{2} w^{2} arctan (w) - \frac{1}{2} (- arctan (w) + w))

Substitute back

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

Simplify

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

= u arctan (u + 1) + 2 arctan (u + 1) - u + 1

Add a constant to the solution

= u arctan (u + 1) + 2 arctan (u + 1) - u + 1 + C

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What is the integral of arctan(sqrt(u+1)) ?
The integral of arctan(sqrt(u+1)) is uarctan(sqrt(u+1))+2arctan(sqrt(u+1))-sqrt(u+1)+C