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

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

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integral from 1 to 2 of (sqrt(x-1))/x

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

2 (- \frac{π}{4} + 1)

Decimal Notation

0.42920 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

\int_{0}^{1} \frac{1}{u ^{2} + 1} d u = \frac{π}{4}

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

= 2 (- \frac{π}{4} + 1)

\int e^{x} \cdot 2 x d x

\int \frac{t ^{2}}{1 + t ^{6}} d t

\frac{w}{w + \frac{c}{w}}

y = x^{2} - 6, at x = a

\frac{x ^{2} - 4}{x - 3}

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