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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

2 (1 - ln (3) + ln (2))

Decimal Notation

1.18906 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

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

= 2 (1 - (ln (3) - ln (2)))

Simplify

= 2 (1 - ln (3) + ln (2))

\int_{1}^{4} 2 x + 1 d x

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

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

\int_{0}^{\frac{π}{2}} cos (x) + sin (x) d x

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

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