What is the integral of (x^3+x^2-x^5)/(x^6) ?

The integral of (x^3+x^2-x^5)/(x^6) is -1/(2x^2)-1/(3x^3)-ln|x|+C

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integral of (x^3+x^2-x^5)/(x^6)

\int \frac{x ^{3} + x ^{2} - x ^{5}}{x ^{6}} d x

- \frac{1}{2 x ^{2}} - \frac{1}{3 x ^{3}} - ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{3} + x ^{2} - x ^{5}}{x ^{6}} d x

Expand

\frac{x ^{3} + x ^{2} - x ^{5}}{x ^{6}} : \frac{1}{x ^{3}} + \frac{1}{x ^{4}} - \frac{1}{x}

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= - \frac{1}{2 x ^{2}} - \frac{1}{3 x ^{3}} - ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{2 x ^{2}} - \frac{1}{3 x ^{3}} - ln ∣ x ∣ + C

\frac{d}{d x} (\frac{1}{4} (x^{4} + 8))

g (x) = 3 x^{4} - 140 x^{3}

\int \frac{1}{t} d t

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

\frac{d}{d x} (cos^{n} (x))

What is the integral of (x^3+x^2-x^5)/(x^6) ?
The integral of (x^3+x^2-x^5)/(x^6) is -1/(2x^2)-1/(3x^3)-ln|x|+C