What is the integral of 5arctan(sqrt(x)) ?

The integral of 5arctan(sqrt(x)) is 10(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C

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integral of 5arctan(sqrt(x))

\int 5 arctan (x) d x

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

Solution steps

\int 5 arctan (x) d x

Take the constant out:

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

= 5 \cdot \int arctan (x) d x

Apply u-substitution

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

= 5 \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

= 5 \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

= 5 \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)

= 5 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

Substitute back

u = x

= 5 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x))

Simplify

5 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : 10 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

= 10 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

Add a constant to the solution

= 10 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

\frac{\partial}{\partial x} (- e^{- 2 t} sin (x))

\int \frac{1}{8} e^{- \frac{1}{8} x} d x

x \to 7 lim (2 x)

\int (12.5 - 0.02 x) d x

\int (cos (5 x))^{2} d x

What is the integral of 5arctan(sqrt(x)) ?
The integral of 5arctan(sqrt(x)) is 10(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C