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

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

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integral of (xsin(x)+cos(x)-1)/(x^2)

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

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

Solution steps

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

Expand

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

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

\int \frac{sin ( x )}{x} d x = Si (x)

\int \frac{cos ( x )}{x ^{2}} d x = - \frac{cos ( x )}{x} - Si (x)

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

= Si (x) - \frac{cos ( x )}{x} - Si (x) - (- \frac{1}{x})

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (ln (x^{2} + 3 x y + 2 y^{2}))

7 x^{2} y + \frac{d y}{d x} = 4 x^{2}

\frac{d}{d x} (y^{2} - 2 x y + 6 x)

k = 0 \sum \infty (1)^{k}

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

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