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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

= 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

= 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

in v erse l a pl a ce \frac{s + 1}{s ( s ^{2} + s + 1 )}

\int_{0}^{1} 2 x - 3 d x

\int_{0}^{1} sin (π x) d x

\frac{\partial}{\partial y} (x e^{x + 2 y})

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