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Integral of 1/(x^2-1)

Step-by-step integral of 1/(x^2-1), with the method named and a graph of the function and its antiderivative.

Kind of integral
1x2−1\frac{1}{x^{2}-1}x2−11​

Antiderivative

∫1x2−1 dx=−ln⁡(∣x+1∣)2+ln⁡(∣x−1∣)2+C\int \frac{1}{x^{2}-1}\,dx=-\frac{\ln\left(\left|x+1\right|\right)}{2}+\frac{\ln\left(\left|x-1\right|\right)}{2}+C∫x2−11​dx=−2ln(∣x+1∣)​+2ln(∣x−1∣)​+C
  • f(x)
  • F(x), with C = 0
Table of values
Graph of 1/(x^2-1): sampled values across the visible range
xf(x)F(x), with C = 0
-60.0285710.16824
-4.80.0453720.21143
-3.60.0836120.28527
-2.40.210080.44365
-1.22.27271.1989
0-10
1.22.2727-1.1989
2.40.21008-0.44365
3.60.083612-0.28527
4.80.045372-0.21143
60.028571-0.16824

Step-by-step working

  1. 1Partial fractions
    1x2−1\frac{1}{x^{2}-1}x2−11​
    ↓
    −12(x+1)+12(x−1)-\frac{1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)}−2(x+1)1​+2(x−1)1​

    Split the denominator into factors and solve for the numerators by matching coefficients.

  2. 2Integrate each term
    ∫1x2−1 dx\int \frac{1}{x^{2}-1}\,dx∫x2−11​dx
    ↓
    −ln⁡(∣x+1∣)2+ln⁡(∣x−1∣)2-\frac{\ln\left(\left|x+1\right|\right)}{2}+\frac{\ln\left(\left|x-1\right|\right)}{2}−2ln(∣x+1∣)​+2ln(∣x−1∣)​

    Linear factors give logarithms (or powers when repeated); an irreducible quadratic gives a logarithm and an arctangent.

  3. 3Add the constant
    −ln⁡(∣x+1∣)2+ln⁡(∣x−1∣)2-\frac{\ln\left(\left|x+1\right|\right)}{2}+\frac{\ln\left(\left|x-1\right|\right)}{2}−2ln(∣x+1∣)​+2ln(∣x−1∣)​
    ↓
    −ln⁡(∣x+1∣)2+ln⁡(∣x−1∣)2+C-\frac{\ln\left(\left|x+1\right|\right)}{2}+\frac{\ln\left(\left|x-1\right|\right)}{2}+C−2ln(∣x+1∣)​+2ln(∣x−1∣)​+C

    Every antiderivative differs by a constant.

Worked solution: integral of 1/(x^2-1)

Worked steps

  1. 1Partial fractions
    1x2−1\frac{1}{x^{2}-1}x2−11​
    ↓
    −12(x+1)+12(x−1)-\frac{1}{2\left(x+1\right)}+\frac{1}{2\left(x-1\right)}−2(x+1)1​+2(x−1)1​

    Split the denominator into factors and solve for the numerators by matching coefficients.

  2. 2Integrate each term
    ∫1x2−1 dx\int \frac{1}{x^{2}-1}\,dx∫x2−11​dx
    ↓
    −ln⁡(∣x+1∣)2+ln⁡(∣x−1∣)2-\frac{\ln\left(\left|x+1\right|\right)}{2}+\frac{\ln\left(\left|x-1\right|\right)}{2}−2ln(∣x+1∣)​+2ln(∣x−1∣)​

    Linear factors give logarithms (or powers when repeated); an irreducible quadratic gives a logarithm and an arctangent.

  3. 3Add the constant
    −ln⁡(∣x+1∣)2+ln⁡(∣x−1∣)2-\frac{\ln\left(\left|x+1\right|\right)}{2}+\frac{\ln\left(\left|x-1\right|\right)}{2}−2ln(∣x+1∣)​+2ln(∣x−1∣)​
    ↓
    −ln⁡(∣x+1∣)2+ln⁡(∣x−1∣)2+C-\frac{\ln\left(\left|x+1\right|\right)}{2}+\frac{\ln\left(\left|x-1\right|\right)}{2}+C−2ln(∣x+1∣)​+2ln(∣x−1∣)​+C

    Every antiderivative differs by a constant.

Answer

∫1x2−1 dx=12ln⁡∣x−1x+1∣+C\int \frac{1}{x^{2}-1}\,dx=\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+C∫x2−11​dx=21​ln​x+1x−1​​+C
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  4. Integral of 1/(x^2-1)

About Integral of 1/(x^2-1)

The integral of 1/(x^2-1) with respect to x is ln(abs((x-1)/(x+1)))/2 + C. Split into partial fractions: 1/(x^2-1) = (1/2)(1/(x-1) - 1/(x+1)). Each piece integrates to a logarithm. It is free, needs no sign-up and runs in your browser, so nothing you enter is uploaded.

The calculator below opens with this problem already worked, and the full steps are written out above. Switch to Definite and give two bounds to see the area, shaded on the graph. For example, from 2 to 3 the area is 0.20273255.

You can always check the answer by differentiating it: the derivative of ln(abs((x-1)/(x+1)))/2 is 1/(x^2-1).

How to use Integral of 1/(x^2-1)

  1. 1Type the function in the Function box, for example x*e^x, or use the key pad. The preview shows how it was read.
  2. 2Choose Indefinite for an antiderivative, or Definite to enter a lower and an upper bound.
  3. 3Write bounds as numbers or constant expressions such as pi/2. Type inf or -inf for an improper integral.
  4. 4Read the result card. A definite integral shows the exact value when one exists and a labelled numeric estimate otherwise.
  5. 5Check the graph: the function is drawn with its area shaded between the bounds, or with its antiderivative for an indefinite integral.
  6. 6Open the steps to see which method was used, and copy the answer with the copy button.

Features

  • Indefinite integrals with the constant of integration.
  • Definite integrals with exact values, including bounds written with pi and with inf or -inf.
  • Step-by-step working that names the rule: power rule, substitution, integration by parts, partial fractions and trigonometric identities.
  • A shaded area under the curve for definite integrals, on a graph you can zoom and drag.
  • A clear label when the result is a numeric estimate instead of an exact value.
  • An honest "no elementary rule found" message instead of a guess.
  • A live preview and a math key pad, so input is easy on a phone.
  • Checks improper integrals and says when one does not settle on a finite value.

The result and the method behind it

Split into partial fractions: 1/(x^2-1) = (1/2)(1/(x-1) - 1/(x+1)). Each piece integrates to a logarithm.

integral 1/(x^2-1) dx = ln(abs((x-1)/(x+1)))/2 + C

Worked examples

  • Integral of 1/(x^2-1) dx → gives ln(abs((x-1)/(x+1)))/2 + C

    Split into partial fractions: 1/(x^2-1) = (1/2)(1/(x-1) - 1/(x+1)). Each piece integrates to a logarithm.

  • Integral of 1/(x^2-1) from 2 to 3 → gives 0.20273255

    The antiderivative at the upper bound minus the antiderivative at the lower bound.

Privacy: how it works

  • The integral is worked out by code running in your browser tab. Your functions are not sent to Ulkaa or to any other server, and there is nothing to sign in to.

Tips

  • Differentiate your answer with the derivative calculator to check it. You should get the original function back.
  • For an area between a curve and the x axis, remember that parts below the axis count as negative in a definite integral.
  • An improper integral with a bound at infinity only has a value if the function dies away fast enough. 1/x^2 does, 1/x does not.
  • Use * between factors when you can, for example x*e^x, and write powers of e as e^x.

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

It is ln(abs((x-1)/(x+1)))/2 + C. Split into partial fractions: 1/(x^2-1) = (1/2)(1/(x-1) - 1/(x+1)). Each piece integrates to a logarithm.

What is the definite integral of 1/(x^2-1) from 2 to 3?

Evaluate the antiderivative ln(abs((x-1)/(x+1)))/2 at the two bounds and subtract. The result is 0.20273255.

Why can I not integrate from 0 to 2 here?

The integrand blows up at x = 1, which lies inside that interval, so the integral is improper and does not converge. The interval from 2 to 3 avoids the problem.

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