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

The integral of ((t^3-8t+1)/((2t)^4)) is 1/16 ln|t|+1/(4t^2)-1/(48t^3)+C

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integral of ((t^3-8t+1)/((2t)^4))

\int (\frac{t ^{3} - 8 t + 1}{( 2 t ) ^{4}}) d t

\frac{1}{16} ln ∣ t ∣ + \frac{1}{4 t ^{2}} - \frac{1}{48 t ^{3}} + C

Solution steps

\int \frac{t ^{3} - 8 t + 1}{( 2 t ) ^{4}} d t

Expand

\frac{t ^{3} - 8 t + 1}{( 2 t ) ^{4}} : \frac{1}{16 t} - \frac{1}{2 t ^{3}} + \frac{1}{16 t ^{4}}

= \int \frac{1}{16 t} - \frac{1}{2 t ^{3}} + \frac{1}{16 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

= \int \frac{1}{16 t} d t - \int \frac{1}{2 t ^{3}} d t + \int \frac{1}{16 t ^{4}} d t

\int \frac{1}{16 t} d t = \frac{1}{16} ln ∣ t ∣

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

\int \frac{1}{16 t ^{4}} d t = - \frac{1}{48 t ^{3}}

= \frac{1}{16} ln ∣ t ∣ - (- \frac{1}{4 t ^{2}}) - \frac{1}{48 t ^{3}}

Simplify

= \frac{1}{16} ln ∣ t ∣ + \frac{1}{4 t ^{2}} - \frac{1}{48 t ^{3}}

Add a constant to the solution

= \frac{1}{16} ln ∣ t ∣ + \frac{1}{4 t ^{2}} - \frac{1}{48 t ^{3}} + C

y^{^{''}} + 36 y = sin (6 t)

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

sin (x) cos^{2} (x), 4 - 4 x, 0.69275, 0.92811

\frac{d y}{d x} + \frac{3 y}{x} + 2 = 3 x, y (1) = 1

\frac{d}{d x} (3 x^{2} + 5 x - 2)

What is the integral of ((t^3-8t+1)/((2t)^4)) ?
The integral of ((t^3-8t+1)/((2t)^4)) is 1/16 ln|t|+1/(4t^2)-1/(48t^3)+C