What is the integral of ((x-4)^2)/(2x^4) ?

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

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integral of ((x-4)^2)/(2x^4)

\int \frac{( x - 4 ) ^{2}}{2 x ^{4}} d x

\frac{1}{2} (- \frac{1}{x} + \frac{4}{x ^{2}} - \frac{16}{3 x ^{3}}) + C

Solution steps

\int \frac{( x - 4 ) ^{2}}{2 x ^{4}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{( x - 4 ) ^{2}}{x ^{4}} d x

Expand

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

= \frac{1}{2} \cdot \int \frac{1}{x ^{2}} - \frac{8}{x ^{3}} + \frac{16}{x ^{4}} d x

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}{2} (\int \frac{1}{x ^{2}} d x - \int \frac{8}{x ^{3}} d x + \int \frac{16}{x ^{4}} d x)

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

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

\int \frac{16}{x ^{4}} d x = - \frac{16}{3 x ^{3}}

= \frac{1}{2} (- \frac{1}{x} - (- \frac{4}{x ^{2}}) - \frac{16}{3 x ^{3}})

Simplify

= \frac{1}{2} (- \frac{1}{x} + \frac{4}{x ^{2}} - \frac{16}{3 x ^{3}})

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{x} + \frac{4}{x ^{2}} - \frac{16}{3 x ^{3}}) + C

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