What is the integral of (2t^3-4t^2+1)/(2t) ?

The integral of (2t^3-4t^2+1)/(2t) is (t^3)/3-t^2+1/2 ln|t|+C

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integral of (2t^3-4t^2+1)/(2t)

\int \frac{2 t ^{3} - 4 t ^{2} + 1}{2 t} d t

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

Solution steps

\int \frac{2 t ^{3} - 4 t ^{2} + 1}{2 t} d t

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{2 t ^{3} - 4 t ^{2} + 1}{t} d t

Expand

\frac{2 t ^{3} - 4 t ^{2} + 1}{t} : 2 t^{2} - 4 t + \frac{1}{t}

= \frac{1}{2} \cdot \int 2 t^{2} - 4 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

= \frac{1}{2} (\int 2 t^{2} d t - \int 4 t d t + \int \frac{1}{t} d t)

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

\int 4 t d t = 2 t^{2}

\int \frac{1}{t} d t = ln ∣ t ∣

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

Simplify

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

= \frac{t ^{3}}{3} - t^{2} + \frac{1}{2} ln ∣ t ∣

Add a constant to the solution

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

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What is the integral of (2t^3-4t^2+1)/(2t) ?
The integral of (2t^3-4t^2+1)/(2t) is (t^3)/3-t^2+1/2 ln|t|+C