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

The integral from 0 to 1 of (sqrt(x-1))/x is 2(-Undefined-i)

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 (sqrt(x-1))/x

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

2 (- Undefined - i)

Solution steps

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

Apply u-substitution

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

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

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= 2 \cdot \int_{i}^{0} - \frac{1}{u ^{2} + 1} + 1 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_{i}^{0} \frac{1}{u ^{2} + 1} d u + \int_{i}^{0} 1 d u)

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

Undefined

\int_{i}^{0} 1 d u = - i

= 2 (- Undefined - i)

\int_{0}^{1} 3 t^{2} sin (t^{3}) d t

\int_{1}^{e} \frac{( ln ( x ^{4} ) )}{x} d x

\int_{1}^{l n (8)} e^{y} (y - 1) d y

\int_{- 2}^{1} 4 x + 8 d x

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

What is the integral from 0 to 1 of (sqrt(x-1))/x ?
The integral from 0 to 1 of (sqrt(x-1))/x is 2(-Undefined-i)