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

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

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

integral

\frac{x ^{2} + 1}{x ^{2} - 1}

- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x + C

Solution steps

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

\frac{x ^{2} + 1}{x ^{2} - 1} = \frac{2}{x ^{2} - 1} + 1

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

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

\int 1 d x = x

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x

Add a constant to the solution

= - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + x + C

\frac{1}{( 1 + x ) ^{0.5}}

2 sin (x) - x^{2} + 1

\frac{( x ^{2} + 4 )}{( x + 3 )}

f (x) = \frac{3 x + 3}{( 3 - 8 x + x ^{2} ) ^{\frac{1}{2}}}

x^{5} (x^{3} + 4)^{\frac{1}{2}}

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