What is the integral of (8-sin(x))/(x^2) ?

The integral of (8-sin(x))/(x^2) is -8/x+(sin(x))/x-\Ci(x)+C

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integral of (8-sin(x))/(x^2)

\int \frac{8 - sin ( x )}{x ^{2}} d x

- \frac{8}{x} + \frac{sin ( x )}{x} - Ci (x) + C

Solution steps

\int \frac{8 - sin ( x )}{x ^{2}} d x

Expand

\frac{8 - sin ( x )}{x ^{2}} : \frac{8}{x ^{2}} - \frac{sin ( x )}{x ^{2}}

= \int \frac{8}{x ^{2}} - \frac{sin ( x )}{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{8}{x ^{2}} d x - \int \frac{sin ( x )}{x ^{2}} d x

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

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

= - \frac{8}{x} - (- \frac{sin ( x )}{x} + Ci (x))

Simplify

= - \frac{8}{x} + \frac{sin ( x )}{x} - Ci (x)

Add a constant to the solution

= - \frac{8}{x} + \frac{sin ( x )}{x} - Ci (x) + C

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What is the integral of (8-sin(x))/(x^2) ?
The integral of (8-sin(x))/(x^2) is -8/x+(sin(x))/x-\Ci(x)+C