Integrating inverse trig and hyperbolic functions

The trick

∫ arsinh x dx = ∫ 1 × arsinh x dx. Take u = arsinh x (so du/dx = 1/sqrt(1 + x²)) and dv/dx = 1 (so v = x):

∫ arsinh x dx = x arsinh x - ∫ x/sqrt(1 + x²) dx = x arsinh x - sqrt(1 + x²) + c.

The family of results

• ∫ arcsin x dx = x arcsin x + sqrt(1 - x²) + c
• ∫ arctan x dx = x arctan x - (1/2) ln(1 + x²) + c
• ∫ arsinh x dx = x arsinh x - sqrt(1 + x²) + c
• ∫ arcosh x dx = x arcosh x - sqrt(x² - 1) + c
• ∫ artanh x dx = x artanh x + (1/2) ln(1 - x²) + c

The second integral in each case is of the form f'(x)/f(x) or f'(x)/sqrt(f(x)).

Exact answers

Use log forms where helpful: arsinh 1 = ln(1 + sqrt(2)), arcosh 2 = ln(2 + sqrt(3)). Check your antiderivative by differentiating it.

Worked example

Find the exact value of ∫ from 0 to 1 of arctan x dx.

1. u = arctan x, dv/dx = 1: [x arctan x] - ∫ x/(1 + x²) dx
2. ∫ x/(1 + x²) dx = (1/2) ln(1 + x²)
3. [x arctan x - (1/2) ln(1 + x²)] from 0 to 1 = π/4 - (1/2) ln 2

Answer: π/4 - (1/2) ln 2, about 0.439

Check yourself

1. Evaluate ∫ from 0 to 1 of arctan x dx, to 3 decimal places.
   Show answer

   0.439. π/4 - (1/2) ln 2 = 0.439.

2. Evaluate ∫ from 0 to 1 of arsinh x dx, to 3 decimal places.
   Show answer

   0.467. [x arsinh x - sqrt(1 + x²)] from 0 to 1 = ln(1 + sqrt(2)) - sqrt(2) + 1 = 0.467.

3. Which is ∫ arcsin x dx?
   • x arcsin x - sqrt(1 - x²) + c
   • 1/sqrt(1 - x²) + c
   • x arcsin x + sqrt(1 - x²) + c
   • x arcsin x + c
   Show answer

   x arcsin x + sqrt(1 - x²) + c. By parts: x arcsin x - ∫ x/sqrt(1 - x²) dx, and ∫ x/sqrt(1 - x²) dx = -sqrt(1 - x²). Subtracting a negative gives +.

4. Evaluate ∫ from 0 to 1/2 of arcsin x dx, to 3 decimal places.
   Show answer

   0.128. [x arcsin x + sqrt(1 - x²)] from 0 to 1/2 = π/12 + sqrt(3)/2 - 1 = 0.128.

5. What is the best first step for ∫ arctan x dx?
   • Substitute x = tan u
   • Use partial fractions
   • Integrate by parts with u = 1
   • Write it as ∫ 1 × arctan x dx and integrate by parts with u = arctan x
   Show answer

   Write it as ∫ 1 × arctan x dx and integrate by parts with u = arctan x. You can differentiate arctan x but you cannot integrate it directly, so it must be u.

6. Evaluate ∫ from 1 to 2 of arcosh x dx, to 3 decimal places.
   Show answer

   0.902. [x arcosh x - sqrt(x² - 1)] from 1 to 2 = 2 ln(2 + sqrt(3)) - sqrt(3) = 0.902.

7. Evaluate ∫ from 0 to 1/2 of artanh x dx, to 3 decimal places.
   Show answer

   0.131. [x artanh x + (1/2) ln(1 - x²)] from 0 to 1/2 = (1/2) artanh(1/2) + (1/2) ln(3/4) = 0.131.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (issue 3), Unit FP3

Last updated 8 November 2026. Spotted a mistake? Tell us.