What is the integral from 1 to 10 of (sqrt(x-1))/x ?

The integral from 1 to 10 of (sqrt(x-1))/x is 2(-arctan(3)+3)

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integral from 1 to 10 of (sqrt(x-1))/x

\int_{1}^{10} \frac{x - 1}{x} d x

2 (- arctan (3) + 3)

Decimal Notation

3.50190 \dots

Solution steps

\int_{1}^{10} \frac{x - 1}{x} d x

Apply u-substitution

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

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

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

= 2 \cdot \int_{0}^{3} - \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}^{3} \frac{1}{u ^{2} + 1} d u + \int_{0}^{3} 1 d u)

\int_{0}^{3} \frac{1}{u ^{2} + 1} d u = arctan (3)

\int_{0}^{3} 1 d u = 3

= 2 (- arctan (3) + 3)

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What is the integral from 1 to 10 of (sqrt(x-1))/x ?
The integral from 1 to 10 of (sqrt(x-1))/x is 2(-arctan(3)+3)