What is the integral of (x^5+1)/(x^6-x^4) ?

The integral of (x^5+1)/(x^6-x^4) is 1/x+1/(3x^3)+ln|x-1|+C

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integral of (x^5+1)/(x^6-x^4)

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

= - (- \frac{1}{x}) - (- \frac{1}{3 x ^{3}}) + ln ∣ x - 1 ∣

Simplify

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

Add a constant to the solution

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

e^{- x} y^{^{'}} = \frac{x}{y}

x (x + 3) d y - y (2 x + 3) d x = 0

(x^{3} + 2)^{^{'}}

\int 2 t - cos (6 t) d t

\int_{0}^{1} \frac{34}{4 y - 1} d y

What is the integral of (x^5+1)/(x^6-x^4) ?
The integral of (x^5+1)/(x^6-x^4) is 1/x+1/(3x^3)+ln|x-1|+C