What is the integral of (x^2sin(x)-2x+3)/(x^2) ?

The integral of (x^2sin(x)-2x+3)/(x^2) is -cos(x)-2ln|x|-3/x+C

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integral of (x^2sin(x)-2x+3)/(x^2)

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

- cos (x) - 2 ln ∣ x ∣ - \frac{3}{x} + C

Solution steps

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

Expand

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

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

\int sin (x) d x = - cos (x)

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= - cos (x) - 2 ln ∣ x ∣ - \frac{3}{x}

Add a constant to the solution

= - cos (x) - 2 ln ∣ x ∣ - \frac{3}{x} + C

\frac{d y}{d t} = - \frac{6}{( 2000 + 6 t )}

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

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

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

\int - 5 tan^{3} (x) d x

What is the integral of (x^2sin(x)-2x+3)/(x^2) ?
The integral of (x^2sin(x)-2x+3)/(x^2) is -cos(x)-2ln|x|-3/x+C