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

The integral of (x^2-x+1)/(x^3) is ln|x|+1/x-1/(2x^2)+C

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

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

ln ∣ x ∣ + \frac{1}{x} - \frac{1}{2 x ^{2}} + C

Solution steps

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

Expand

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Simplify

= ln ∣ x ∣ + \frac{1}{x} - \frac{1}{2 x ^{2}}

Add a constant to the solution

= ln ∣ x ∣ + \frac{1}{x} - \frac{1}{2 x ^{2}} + C

\frac{d}{d x} (- 5 sin (x) cos (x))

f (x) = 8 x^{7} e^{x} + e^{x} x^{8}

\frac{d}{d x} (2 sinh (2 x))

n = 1 \sum \infty n^{3} sin (\frac{1}{n})

f (x) = \frac{9}{x ^{2}}

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