What is the integral of (1-3t)/(t^2) ?

The integral of (1-3t)/(t^2) is -1/t-3ln|t|+C

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integral of (1-3t)/(t^2)

\int \frac{1 - 3 t}{t ^{2}} d t

- \frac{1}{t} - 3 ln ∣ t ∣ + C

Solution steps

\int \frac{1 - 3 t}{t ^{2}} d t

Expand

\frac{1 - 3 t}{t ^{2}} : \frac{1}{t ^{2}} - \frac{3}{t}

= \int \frac{1}{t ^{2}} - \frac{3}{t} d t

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}{t ^{2}} d t - \int \frac{3}{t} d t

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

\int \frac{3}{t} d t = 3 ln ∣ t ∣

= - \frac{1}{t} - 3 ln ∣ t ∣

Add a constant to the solution

= - \frac{1}{t} - 3 ln ∣ t ∣ + C

f (x) = \frac{4 x ^{2} - 7 x + 5}{2 x + 3}

f (x) = 15 e^{x^{2}}

f (x) = 3 x^{2} + 12

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

\frac{\partial}{\partial x} (ln (sin (x y)))

What is the integral of (1-3t)/(t^2) ?
The integral of (1-3t)/(t^2) is -1/t-3ln|t|+C