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

The integral of 1/((x+1)sqrt(x)) is 2arctan(sqrt(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

2 arctan (x) + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 2 arctan (u)

Substitute back

u = x

= 2 arctan (x)

Add a constant to the solution

= 2 arctan (x) + C

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