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

The integral from 1 to 4 of (sqrt(x-1))/x is 2(-pi/3+sqrt(3))

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integral from 1 to 4 of (sqrt(x-1))/x

\int_{1}^{4} \frac{x - 1}{x} d x

2 (- \frac{π}{3} + 3)

Decimal

1.36970 \dots

Solution steps

\int_{1}^{4} \frac{x - 1}{x} d x

Apply u-substitution

: \int_{0}^{3} \frac{2 u ^{2}}{u ^{2} + 1} d u

= \int_{0}^{3} \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}^{3} \frac{u ^{2}}{u ^{2} + 1} d u

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

= 2 \cdot \int_{0}^{3} - \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}^{3} \frac{1}{u ^{2} + 1} d u + \int_{0}^{3} 1 d u)

\int_{0}^{3} \frac{1}{u ^{2} + 1} d u = \frac{π}{3}

\int_{0}^{3} 1 d u = 3

= 2 (- \frac{π}{3} + 3)

\int_{2}^{5} 2^{4 x} d x

\int_{- 1}^{8} \frac{1}{x ^{3}} d x

\int_{0}^{2} \frac{x + 1}{x ^{2} + 1} d x

\int_{0}^{2} (2 x - 9) (8 x^{2} + 6) d x

\int_{0}^{1} 2 + 3 x d x

What is the integral from 1 to 4 of (sqrt(x-1))/x ?
The integral from 1 to 4 of (sqrt(x-1))/x is 2(-pi/3+sqrt(3))