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

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

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

\int arctan (2 x + 1) d x

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

Solution steps

\int arctan (2 x + 1) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

= \frac{1}{2} \cdot \int arctan (v) \cdot 2 v d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int arctan (v) v d v

Apply Integration By Parts

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

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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