What is the integral of (t^3-8t+1)/(8t^4) ?

The integral of (t^3-8t+1)/(8t^4) is 1/8 (ln|t|+4/(t^2)-1/(3t^3))+C

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integral of (t^3-8t+1)/(8t^4)

\int \frac{t ^{3} - 8 t + 1}{8 t ^{4}} d t

\frac{1}{8} (ln ∣ t ∣ + \frac{4}{t ^{2}} - \frac{1}{3 t ^{3}}) + C

Solution steps

\int \frac{t ^{3} - 8 t + 1}{8 t ^{4}} d t

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{t ^{3} - 8 t + 1}{t ^{4}} d t

Expand

\frac{t ^{3} - 8 t + 1}{t ^{4}} : \frac{1}{t} - \frac{8}{t ^{3}} + \frac{1}{t ^{4}}

= \frac{1}{8} \cdot \int \frac{1}{t} - \frac{8}{t ^{3}} + \frac{1}{t ^{4}} 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}{8} (\int \frac{1}{t} d t - \int \frac{8}{t ^{3}} d t + \int \frac{1}{t ^{4}} d t)

\int \frac{1}{t} d t = ln ∣ t ∣

\int \frac{8}{t ^{3}} d t = - \frac{4}{t ^{2}}

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

= \frac{1}{8} (ln ∣ t ∣ - (- \frac{4}{t ^{2}}) - \frac{1}{3 t ^{3}})

Simplify

= \frac{1}{8} (ln ∣ t ∣ + \frac{4}{t ^{2}} - \frac{1}{3 t ^{3}})

Add a constant to the solution

= \frac{1}{8} (ln ∣ t ∣ + \frac{4}{t ^{2}} - \frac{1}{3 t ^{3}}) + C

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What is the integral of (t^3-8t+1)/(8t^4) ?
The integral of (t^3-8t+1)/(8t^4) is 1/8 (ln|t|+4/(t^2)-1/(3t^3))+C