What is the integral of (1-1/x)^3*1/(x^2) ?

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

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integral of (1-1/x)^3*1/(x^2)

\int (1 - \frac{1}{x})^{3} \cdot \frac{1}{x ^{2}} d x

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

Solution steps

\int (1 - \frac{1}{x})^{3} \frac{1}{x ^{2}} d x

Expand

(1 - \frac{1}{x})^{3} \frac{1}{x ^{2}} : \frac{1}{x ^{2}} - \frac{3}{x ^{3}} + \frac{3}{x ^{4}} - \frac{1}{x ^{5}}

= \int \frac{1}{x ^{2}} - \frac{3}{x ^{3}} + \frac{3}{x ^{4}} - \frac{1}{x ^{5}} 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

= \int \frac{1}{x ^{2}} d x - \int \frac{3}{x ^{3}} d x + \int \frac{3}{x ^{4}} d x - \int \frac{1}{x ^{5}} d x

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

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

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

\int \frac{1}{x ^{5}} d x = - \frac{1}{4 x ^{4}}

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

Simplify

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

Add a constant to the solution

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

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