What is the integral of 24arctan(sqrt(x)) ?

The integral of 24arctan(sqrt(x)) is 48(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C

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integral of 24arctan(sqrt(x))

\int 24 arctan (x) d x

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

Solution steps

\int 24 arctan (x) d x

Take the constant out:

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

= 24 \cdot \int arctan (x) d x

Apply u-substitution

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

= 24 \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

= 24 \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

= 24 \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)

= 24 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

Substitute back

u = x

= 24 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x))

Simplify

24 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : 48 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

= 48 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

Add a constant to the solution

= 48 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

\int sec^{6} (2 x) tan (2 x) d x

cos (x) \frac{d y}{d x} - sin (x) y = 0

\frac{\partial}{\partial x} (\frac{x - y}{x y + 2})

x \to \infty lim (ln (1))

y^{^{'}} = 8 x y - 2 x, y (0) = 16

What is the integral of 24arctan(sqrt(x)) ?
The integral of 24arctan(sqrt(x)) is 48(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C