What is the integral of (x^2-10x+4)/(x^3) ?

The integral of (x^2-10x+4)/(x^3) is ln|x|+(10)/x-2/(x^2)+C

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integral of (x^2-10x+4)/(x^3)

\int \frac{x ^{2} - 10 x + 4}{x ^{3}} d x

ln ∣ x ∣ + \frac{10}{x} - \frac{2}{x ^{2}} + C

Solution steps

\int \frac{x ^{2} - 10 x + 4}{x ^{3}} d x

Expand

\frac{x ^{2} - 10 x + 4}{x ^{3}} : \frac{1}{x} - \frac{10}{x ^{2}} + \frac{4}{x ^{3}}

= \int \frac{1}{x} - \frac{10}{x ^{2}} + \frac{4}{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{10}{x ^{2}} d x + \int \frac{4}{x ^{3}} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{10}{x ^{2}} d x = - \frac{10}{x}

\int \frac{4}{x ^{3}} d x = - \frac{2}{x ^{2}}

= ln ∣ x ∣ - (- \frac{10}{x}) - \frac{2}{x ^{2}}

Simplify

= ln ∣ x ∣ + \frac{10}{x} - \frac{2}{x ^{2}}

Add a constant to the solution

= ln ∣ x ∣ + \frac{10}{x} - \frac{2}{x ^{2}} + C

\frac{d}{d x} (2 - e^{- x})

n = 0 \sum \infty \frac{1}{2 ^{2 n + 1}}

x \to \infty lim (1)

\int \frac{2 t}{t ^{2} + 1} d t

f (x) = ∣ 2 x - 3 ∣

What is the integral of (x^2-10x+4)/(x^3) ?
The integral of (x^2-10x+4)/(x^3) is ln|x|+(10)/x-2/(x^2)+C