What is the integral from 0 to 4 of 1/(sqrt(x)+1) ?

The integral from 0 to 4 of 1/(sqrt(x)+1) is 2(2-ln(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 4 of 1/(sqrt(x)+1)

\int_{0}^{4} \frac{1}{x + 1} d x

2 (2 - ln (3))

Decimal

1.80277 \dots

Solution steps

\int_{0}^{4} \frac{1}{x + 1} d x

Apply u-substitution

= \int_{1}^{3} \frac{2 ( u - 1 )}{u} d u

Take the constant out:

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

= 2 \cdot \int_{1}^{3} \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 2 \cdot \int_{1}^{3} 1 - \frac{1}{u} 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_{1}^{3} 1 d u - \int_{1}^{3} \frac{1}{u} d u)

\int_{1}^{3} 1 d u = 2

\int_{1}^{3} \frac{1}{u} d u = ln (3)

= 2 (2 - ln (3))

What is the integral from 0 to 4 of 1/(sqrt(x)+1) ?
The integral from 0 to 4 of 1/(sqrt(x)+1) is 2(2-ln(3))