What is the integral of arctan(1+sqrt(x)) ?

The integral of arctan(1+sqrt(x)) is xarctan(1+sqrt(x))+ln|x+2sqrt(x)+2|-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 arctan(1+sqrt(x))

\int arctan (1 + x) d x

x arctan (1 + x) + ln x + 2 x + 2 - x + C

Solution steps

\int arctan (1 + x) d x

Apply u-substitution

: \int arctan (1 + u) \cdot 2 u d u

= \int arctan (1 + u) \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int arctan (1 + u) u d u

Apply Integration By Parts

: \frac{1}{2} u^{2} arctan (1 + u) - \int \frac{u ^{2}}{2 ( u ^{2} + 2 u + 2 )} d u

= 2 (\frac{1}{2} u^{2} arctan (1 + u) - \int \frac{u ^{2}}{2 ( u ^{2} + 2 u + 2 )} d u)

\int \frac{u ^{2}}{2 ( u ^{2} + 2 u + 2 )} d u = \frac{1}{2} (- ln u^{2} + 2 u + 2 + u)

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

Substitute back

u = x

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

Simplify

2 (\frac{1}{2} (x)^{2} arctan (1 + x) - \frac{1}{2} (- ln (x)^{2} + 2 x + 2 + x)) : x arctan (1 + x) + ln x + 2 x + 2 - x

= x arctan (1 + x) + ln x + 2 x + 2 - x

Add a constant to the solution

= x arctan (1 + x) + ln x + 2 x + 2 - x + C

\int 7 x (x^{2} + 8)^{2} d x

\int \frac{r}{( 5 - 4 r - r ^{2} ) ^{\frac{3}{2}}} d r

3 x

\int \frac{x ^{4}}{x ^{5} - 6} d x

\int 2 cos (2 x + 90) d x

What is the integral of arctan(1+sqrt(x)) ?
The integral of arctan(1+sqrt(x)) is xarctan(1+sqrt(x))+ln|x+2sqrt(x)+2|-sqrt(x)+C