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Popular >

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

Solution

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

Solution

\frac{x ^{2}}{2} - \frac{1}{4} ln x^{2} + 1 + \frac{1}{4} ln x^{2} - 1 + \frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

\frac{x ^{5} - 1}{x ^{4} - 1} : \frac{x ^{5}}{x ^{4} - 1} - \frac{1}{x ^{4} - 1}

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

\int \frac{x ^{5}}{x ^{4} - 1} d x = \frac{x ^{2}}{2} - \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1)

\int \frac{1}{x ^{4} - 1} d x = - \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣

= \frac{x ^{2}}{2} - \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1) - (- \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣)

Simplify

\frac{x ^{2}}{2} - \frac{1}{4} (ln x^{2} + 1 - ln x^{2} - 1) - (- \frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) : \frac{x ^{2}}{2} - \frac{1}{4} ln x^{2} + 1 + \frac{1}{4} ln x^{2} - 1 + \frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣

= \frac{x ^{2}}{2} - \frac{1}{4} ln x^{2} + 1 + \frac{1}{4} ln x^{2} - 1 + \frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{1}{4} ln x^{2} + 1 + \frac{1}{4} ln x^{2} - 1 + \frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ + C

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

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

Frequently Asked Questions (FAQ)

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