What is the integral of 33arctan(sqrt(x)) ?

The integral of 33arctan(sqrt(x)) is 66(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C

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integral of 33arctan(sqrt(x))

\int 33 arctan (x) d x

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

Solution steps

\int 33 arctan (x) d x

Take the constant out:

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

= 33 \cdot \int arctan (x) d x

Apply u-substitution

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

= 33 \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

= 33 \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

= 33 \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)

= 33 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

Substitute back

u = x

= 33 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x))

Simplify

33 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : 66 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

= 66 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

Add a constant to the solution

= 66 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

x \to 0 - lim (\frac{e ^{\frac{2}{x}} \cdot sin ( π ( x ) )}{x + 2} + x)

x \to 0 lim (\frac{8 sin ( 0 )}{0})

x + y = 3, y^{2} = 4 x, y = 0, 1

\int (sin^{2} (x) + cos (x))^{2} d x

f (x) = x - 2 sin (x)

What is the integral of 33arctan(sqrt(x)) ?
The integral of 33arctan(sqrt(x)) is 66(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C