What is the integral of ((x^5+2))/(x^2-1) ?

The integral of ((x^5+2))/(x^2-1) is (x^4)/4+(x^2)/2+1/2 ln|x^2-1|-ln|x+1|+ln|x-1|+C

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integral of ((x^5+2))/(x^2-1)

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

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

Solution steps

\int \frac{x ^{5} + 2}{x ^{2} - 1} d x

Long division

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

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

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

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

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

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

Add a constant to the solution

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

\int sin (16 x) cos (13 x) d x

\int_{0}^{1} 13 e^{x^{2}} d x

x \to 4 lim (- x + 4)

\int (4 e^{x} + 12 x^{\frac{3}{2}} + 4 x^{\frac{1}{2}}) d x

x^{2} - 4, at x = 3

What is the integral of ((x^5+2))/(x^2-1) ?
The integral of ((x^5+2))/(x^2-1) is (x^4)/4+(x^2)/2+1/2 ln|x^2-1|-ln|x+1|+ln|x-1|+C