What is the integral of 14arctan(sqrt(x)) ?

The integral of 14arctan(sqrt(x)) is 28(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

Popular >

integral of 14arctan(sqrt(x))

\int 14 arctan (x) d x

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

Solution steps

\int 14 arctan (x) d x

Take the constant out:

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

= 14 \cdot \int arctan (x) d x

Apply u-substitution

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

= 14 \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

= 14 \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

= 14 \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)

= 14 \cdot 2 (\frac{1}{2} u^{2} arctan (u) - \frac{1}{2} (- arctan (u) + u))

Substitute back

u = x

= 14 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x))

Simplify

14 \cdot 2 (\frac{1}{2} (x)^{2} arctan (x) - \frac{1}{2} (- arctan (x) + x)) : 28 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

= 28 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x))

Add a constant to the solution

= 28 (\frac{1}{2} x arctan (x) - \frac{1}{2} (- arctan (x) + x)) + C

\frac{d x}{d t} + t x^{3} + \frac{x}{t} = 0

\int_{t - 1}^{2} - t + 2 d t

f (x) = (2 x + 3)^{3}

\frac{d}{d x} (1 + 3 x)

What is the integral of 14arctan(sqrt(x)) ?
The integral of 14arctan(sqrt(x)) is 28(1/2 xarctan(sqrt(x))-1/2 (-arctan(sqrt(x))+sqrt(x)))+C