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

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

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

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

2 (- \frac{π}{4} + 1)

Decimal

0.42920 \dots

Solution steps

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

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)

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