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

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

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

integral

\frac{( 2 t ^{2} - 1 )}{t}

t^{2} - ln ∣ t ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{t} d t = ln ∣ t ∣

= t^{2} - ln ∣ t ∣

Add a constant to the solution

= t^{2} - ln ∣ t ∣ + C

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