What is the integral of (sin(1/x)-1/x+1) 1/(x^2) ?

The integral of (sin(1/x)-1/x+1) 1/(x^2) is cos(1/x)+1/(2x^2)-1/x+C

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integral of (sin(1/x)-1/x+1) 1/(x^2)

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

cos (\frac{1}{x}) + \frac{1}{2 x ^{2}} - \frac{1}{x} + C

Solution steps

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

Simplify

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

= \int \frac{x sin ( \frac{1}{x} ) - 1 + x}{x ^{3}} d x

Expand

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

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

\int \frac{sin ( \frac{1}{x} )}{x ^{2}} d x = cos (\frac{1}{x})

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

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

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

Simplify

= cos (\frac{1}{x}) + \frac{1}{2 x ^{2}} - \frac{1}{x}

Add a constant to the solution

= cos (\frac{1}{x}) + \frac{1}{2 x ^{2}} - \frac{1}{x} + C

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What is the integral of (sin(1/x)-1/x+1) 1/(x^2) ?
The integral of (sin(1/x)-1/x+1) 1/(x^2) is cos(1/x)+1/(2x^2)-1/x+C