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

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

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

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

- \frac{π}{4} + 1

Decimal

0.21460 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= \frac{1}{2} \cdot \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

= \frac{1}{2} \cdot 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

= \frac{1}{2} \cdot 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

= \frac{1}{2} \cdot 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

= \frac{1}{2} \cdot 2 (- \frac{π}{4} + 1)

Simplify

\frac{1}{2} \cdot 2 (- \frac{π}{4} + 1) : - \frac{π}{4} + 1

= - \frac{π}{4} + 1

\frac{\partial}{\partial y} (sin (\frac{5 x}{y}))

y = \frac{4}{x}, y = x, y = 4

\int \frac{x}{3 x - 4} d x

\int 5 x^{2} (x^{3} + 1)^{10} d x

f (x) = (5 x^{3} + 4) (3 x^{7} - 5)

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