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

The integral of (sqrt(x))/(1+x) is 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 (sqrt(x))/(1+x)

\int \frac{x}{1 + x} d x

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

Solution steps

\int \frac{x}{1 + x} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{1 + u ^{2}} d u

= \int \frac{2 u ^{2}}{1 + u ^{2}} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{1 + u ^{2}} d u

\frac{u ^{2}}{1 + u ^{2}} = - \frac{1}{1 + u ^{2}} + 1

= 2 \cdot \int - \frac{1}{1 + u ^{2}} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d u = u

= 2 (- arctan (u) + u)

Substitute back

u = x

= 2 (- arctan (x) + x)

Add a constant to the solution

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

\int \frac{3 ( x + 5 )}{2 x + 15} d x

\int_{0}^{π} 6 sin (x) + sin (6 x) d x

x \to 4 + lim (\frac{6}{x - 4})

f (x) = x^{2} - 4 x - 5

\int (9 + x^{5})^{7} x^{4} d x

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