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

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

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

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

2 (- arctan (2) + 2)

Decimal

1.78570 \dots

Solution steps

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

Apply u-substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

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

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}{v ^{2} + 1} d v + \int_{0}^{2} 1 d v)

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

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

= 2 (- arctan (2) + 2)

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