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

The integral from 0 to 1 of x/(1+sqrt(x)) is 2(-ln(2)+5/6)

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 x/(1+sqrt(x))

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

2 (- ln (2) + \frac{5}{6})

Decimal

0.28037 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{2 u ^{3}}{1 + u} d u

= \int_{0}^{1} \frac{2 u ^{3}}{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_{0}^{1} \frac{u ^{3}}{1 + u} d u

Apply u-substitution

: \int_{1}^{2} \frac{( v - 1 ) ^{3}}{v} d v

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

Expand

\frac{( v - 1 ) ^{3}}{v} : v^{2} - 3 v + 3 - \frac{1}{v}

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

\int_{1}^{2} v^{2} d v = \frac{7}{3}

\int_{1}^{2} 3 v d v = \frac{9}{2}

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

\int_{1}^{2} \frac{1}{v} d v = ln (2)

= 2 (\frac{7}{3} - \frac{9}{2} + 3 - ln (2))

Combine the fractions

3 + \frac{7}{3} - \frac{9}{2} : \frac{5}{6}

= 2 (- ln (2) + \frac{5}{6})

\int_{0}^{1 - y} 6 y d y

\int_{0}^{y} 3 y d y

\int_{2 + c o s (x)}^{x^{2} + 1} \frac{1}{t} d t

\int_{0}^{1} x^{4} (5 + 3 x^{5})^{4} d x

\int_{0}^{1} - 1 - x^{2} d x

What is the integral from 0 to 1 of x/(1+sqrt(x)) ?
The integral from 0 to 1 of x/(1+sqrt(x)) is 2(-ln(2)+5/6)