What is the integral of (u-1)/(u^2) ?

The integral of (u-1)/(u^2) is ln|u|+1/u+C

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integral of (u-1)/(u^2)

\int \frac{u - 1}{u ^{2}} d u

ln ∣ u ∣ + \frac{1}{u} + C

Solution steps

\int \frac{u - 1}{u ^{2}} d u

Expand

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

= \int \frac{1}{u} - \frac{1}{u ^{2}} 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 \frac{1}{u} d u - \int \frac{1}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

= ln ∣ u ∣ - (- \frac{1}{u})

Simplify

= ln ∣ u ∣ + \frac{1}{u}

Add a constant to the solution

= ln ∣ u ∣ + \frac{1}{u} + C

\int_{0}^{\infty} x e^{- x^{2}} d x

\frac{\partial}{\partial y} (\frac{1}{x - y})

2

f (x) = \frac{9}{x ^{2} + 5}

cos^{2} (x), 0

What is the integral of (u-1)/(u^2) ?
The integral of (u-1)/(u^2) is ln|u|+1/u+C