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

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

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

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

- \frac{1}{y} - ln ∣ y ∣ + C

Solution steps

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

Expand

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

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

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

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

\int \frac{1}{y} d y = ln ∣ y ∣

= - \frac{1}{y} - ln ∣ y ∣

Add a constant to the solution

= - \frac{1}{y} - ln ∣ y ∣ + C

f (θ) = θ sin (θ)

\frac{s + 5}{( s + 10 ) ^{2}}

\frac{d ^{5}}{d x ^{5}} (x)

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

f (x) = (9 x^{6} + 8 x^{3})^{4}

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