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

The integral of 2arctan(sqrt(x)) is 4(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C

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integral of 2arctan(sqrt(x))

\int 2 arctan (x) d x

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

Solution steps

\int 2 arctan (x) d x

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = x

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

Simplify

2 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : 4 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

= 4 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

Add a constant to the solution

= 4 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

f (x) = x^{5} + 4, g (x) = x + 4

f (x) = - \frac{1}{2 1 - x}, at x = - 8

\int \frac{1}{u 7 - u ^{2}} d u

f^{^{'}} (x) = \frac{8 x ^{5} - 6 x ^{9}}{x ^{4}}

\frac{d}{d x} (ln (1 + e^{2 x}))

What is the integral of 2arctan(sqrt(x)) ?
The integral of 2arctan(sqrt(x)) is 4(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C