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Popular >

integral of (u-1)/u

Solution

\int \frac{u - 1}{u} d u

Solution

u - ln ∣ u ∣ + C

Solution steps

\int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \int 1 - \frac{1}{u} 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

= \int 1 d u - \int \frac{1}{u} d u

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u - ln ∣ u ∣

Add a constant to the solution

= u - ln ∣ u ∣ + C

Popular Examples

area y= 1/x ,y= 1/3 ,1,3 area

y = \frac{1}{x}, y = \frac{1}{3}, 1, 3

integral of 1/(x^2+2x-3)

\int \frac{1}{x ^{2} + 2 x - 3} d x

derivative of (x^6/6)

\frac{d}{d x} (\frac{x ^{6}}{6})

parity y=cot^2(sin(θ))parity

y = cot^{2} (sin (θ))

derivative of (x^2/(2x+1))

\frac{d}{d x} (\frac{x ^{2}}{2 x + 1})

Frequently Asked Questions (FAQ)

What is the integral of (u-1)/u ?
The integral of (u-1)/u is u-ln|u|+C