What is the integral of (x^3-6x+5)/(x^4) ?

The integral of (x^3-6x+5)/(x^4) is ln|x|+3/(x^2)-5/(3x^3)+C

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integral of (x^3-6x+5)/(x^4)

\int \frac{x ^{3} - 6 x + 5}{x ^{4}} d x

ln ∣ x ∣ + \frac{3}{x ^{2}} - \frac{5}{3 x ^{3}} + C

Solution steps

\int \frac{x ^{3} - 6 x + 5}{x ^{4}} d x

Expand

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

= \int \frac{1}{x} - \frac{6}{x ^{3}} + \frac{5}{x ^{4}} 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{6}{x ^{3}} d x + \int \frac{5}{x ^{4}} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{5}{x ^{4}} d x = - \frac{5}{3 x ^{3}}

= ln ∣ x ∣ - (- \frac{3}{x ^{2}}) - \frac{5}{3 x ^{3}}

Simplify

= ln ∣ x ∣ + \frac{3}{x ^{2}} - \frac{5}{3 x ^{3}}

Add a constant to the solution

= ln ∣ x ∣ + \frac{3}{x ^{2}} - \frac{5}{3 x ^{3}} + C

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

\int \frac{1}{cos ^{6} ( 4 x )} d x

\int 6 - x - x^{2} d x

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

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

What is the integral of (x^3-6x+5)/(x^4) ?
The integral of (x^3-6x+5)/(x^4) is ln|x|+3/(x^2)-5/(3x^3)+C