What is the integral from 0 to 1 of 1/(1+2sqrt(x)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 1/(1+2sqrt(x))

\int_{0}^{1} \frac{1}{1 + 2 x} d x

1 - \frac{1}{2} ln (3)

Decimal

0.45069 \dots

Solution steps

\int_{0}^{1} \frac{1}{1 + 2 x} d x

Apply u-substitution

: \int_{0}^{1} \frac{2 u}{1 + 2 u} d u

= \int_{0}^{1} \frac{2 u}{1 + 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_{0}^{1} \frac{u}{1 + 2 u} d u

Apply u-substitution

: \int_{1}^{3} \frac{v - 1}{4 v} d v

= 2 \cdot \int_{1}^{3} \frac{v - 1}{4 v} d v

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int_{1}^{3} \frac{v - 1}{v} d v

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

= 2 \cdot \frac{1}{4} \cdot \int_{1}^{3} 1 - \frac{1}{v} d v

Apply the Sum Rule:

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

= 2 \cdot \frac{1}{4} (\int_{1}^{3} 1 d v - \int_{1}^{3} \frac{1}{v} d v)

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

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

= 2 \cdot \frac{1}{4} (2 - ln (3))

Simplify

2 \cdot \frac{1}{4} (2 - ln (3)) : 1 - \frac{1}{2} ln (3)

= 1 - \frac{1}{2} ln (3)

\int_{0}^{2} x sin (\frac{nπ}{2} x) d x

\int_{1}^{2} 8 x^{3} d x

\int_{e}^{e^{2}} \frac{cos ( ln ( x ) )}{x} d x

\int_{- 1}^{4} 4 x (1 - x^{2}) d x

\int_{0}^{1} 4 y^{2} + 1 d y

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