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 = ln (\frac{3 + e ^{x}}{3 - e ^{x}})

f (x) = (ln (x))^{3}, at x = 9

x \to 0 lim (\frac{2 - 2 cos ( x )}{x})

\int_{a}^{b} x d x

y^{2} - 2 x = 9, x - y = 3 [- 4.5, 8]

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