What is the integral from 1 to 5 of (sqrt(x-1))/x ?

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

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integral from 1 to 5 of (sqrt(x-1))/x

\int_{1}^{5} \frac{x - 1}{x} d x

2 (- arctan (2) + 2)

Decimal

1.78570 \dots

Solution steps

\int_{1}^{5} \frac{x - 1}{x} d x

Apply u-substitution

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

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

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

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

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

\int_{0}^{2} 1 d u = 2

= 2 (- arctan (2) + 2)

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