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

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

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

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

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

Solution steps

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

Expand

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

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

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

Simplify

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

Add a constant to the solution

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

f (x) = - 3 x^{2} + 2, at x = 1

f (x) = \frac{4 x ^{3}}{5 - 2 x}

x \to 6 + lim (\frac{4}{x - 6})

f (x) = (3 x^{6} + 4 x^{3})^{4}

\int \frac{1}{u} d u

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