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

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

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

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

- π + 4

Decimal

0.85840 \dots

Solution steps

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

Take the constant out:

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

= 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

= 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

= 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

= 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

= 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

= 2 \cdot 2 (- \frac{π}{4} + 1)

Simplify

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

= - π + 4

t (\frac{d y}{d t}) + 5 y = 4 t, y (1) = 2

x^{- \frac{3}{2}}

x \to + 5 lim (x^{2} - 2 x + 1)

\int \frac{2 + e ^{x}}{2 - e ^{x}} d x

\frac{\partial}{\partial x} (ln (e^{x y} cos (x)))

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