What is the integral of (3t^4-t^3+6t^2)/(t^4) ?

The integral of (3t^4-t^3+6t^2)/(t^4) is 3t-ln|t|-6/t+C

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integral of (3t^4-t^3+6t^2)/(t^4)

\int \frac{3 t ^{4} - t ^{3} + 6 t ^{2}}{t ^{4}} d t

3 t - ln ∣ t ∣ - \frac{6}{t} + C

Solution steps

\int \frac{3 t ^{4} - t ^{3} + 6 t ^{2}}{t ^{4}} d t

Expand

\frac{3 t ^{4} - t ^{3} + 6 t ^{2}}{t ^{4}} : 3 - \frac{1}{t} + \frac{6}{t ^{2}}

= \int 3 - \frac{1}{t} + \frac{6}{t ^{2}} 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 3 d t - \int \frac{1}{t} d t + \int \frac{6}{t ^{2}} d t

\int 3 d t = 3 t

\int \frac{1}{t} d t = ln ∣ t ∣

\int \frac{6}{t ^{2}} d t = - \frac{6}{t}

= 3 t - ln ∣ t ∣ - \frac{6}{t}

Add a constant to the solution

= 3 t - ln ∣ t ∣ - \frac{6}{t} + C

csc (y) d x + sec^{2} (x) d y = 0

\int e t d t

x \to 0 lim (\frac{∣ x ∣}{x ^{2}})

\int \frac{1}{a + b x} d x

\int \frac{1}{7 + x ^{2}} d x

What is the integral of (3t^4-t^3+6t^2)/(t^4) ?
The integral of (3t^4-t^3+6t^2)/(t^4) is 3t-ln|t|-6/t+C