What is the integral of 34arctan(sqrt(x)) ?

The integral of 34arctan(sqrt(x)) is 68(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C

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integral of 34arctan(sqrt(x))

\int 34 arctan (x) d x

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

Solution steps

\int 34 arctan (x) d x

Take the constant out:

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

= 34 \cdot \int arctan (x) d x

Apply u-substitution

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

= 34 \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

= 34 \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

= 34 \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)

= 34 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

Substitute back

u = x

= 34 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x))

Simplify

34 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : 68 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

= 68 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

Add a constant to the solution

= 68 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

x \to 2 lim ((x + 2) \cdot (2 x - 1))

3 + \frac{1}{5} sin (3 π t)

ln (x + 1), 1

\frac{\partial}{\partial y} (x^{2} y cos (\frac{z}{t}))

y^{^{'}} - \frac{8}{x} y = (\frac{y}{x ^{6}})^{3}

What is the integral of 34arctan(sqrt(x)) ?
The integral of 34arctan(sqrt(x)) is 68(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C