What is the integral of (t^2-3)/(5t) ?

The integral of (t^2-3)/(5t) is 1/5 ((t^2)/2-3ln|t|)+C

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integral of (t^2-3)/(5t)

\int \frac{t ^{2} - 3}{5 t} d t

\frac{1}{5} (\frac{t ^{2}}{2} - 3 ln ∣ t ∣) + C

Solution steps

\int \frac{t ^{2} - 3}{5 t} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int \frac{t ^{2} - 3}{t} d t

Expand

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

= \frac{1}{5} \cdot \int t - \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

= \frac{1}{5} (\int t d t - \int \frac{3}{t} d t)

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

\int \frac{3}{t} d t = 3 ln ∣ t ∣

= \frac{1}{5} (\frac{t ^{2}}{2} - 3 ln ∣ t ∣)

Add a constant to the solution

= \frac{1}{5} (\frac{t ^{2}}{2} - 3 ln ∣ t ∣) + C

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What is the integral of (t^2-3)/(5t) ?
The integral of (t^2-3)/(5t) is 1/5 ((t^2)/2-3ln|t|)+C