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Integrals that give inverse functions

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

Integrals of 1/(a² + x²), 1/sqrt(a² - x²) and 1/sqrt(x² ± a²) give inverse trigonometric and inverse hyperbolic functions. With completing the square, they cover a whole family of integrals.

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The standard results

These are in the formula booklet; be ready to prove them by substitution.

Integral	Result	Substitution to prove it
∫ 1/sqrt(a² - x²) dx	arcsin(x/a) + c, \|x\| < a	x = a sin u
∫ 1/(a² + x²) dx	(1/a) arctan(x/a) + c	x = a tan u
∫ 1/sqrt(x² + a²) dx	arsinh(x/a) or ln(x + sqrt(x² + a²)) + c	x = a sinh u
∫ 1/sqrt(x² - a²) dx	arcosh(x/a) or ln(x + sqrt(x² - a²)) + c, x > a	x = a cosh u
∫ 1/(a² - x²) dx	(1/(2a)) ln((a + x)/(a - x)) = (1/a) artanh(x/a) + c, \|x\| < a	partial fractions

Proof by substitution

∫ 1/sqrt(x² + a²) dx with x = a sinh u, dx = a cosh u du:

sqrt(a² sinh² u + a²) = a cosh u, so the integral is ∫ a cosh u/(a cosh u) du = u + c = arsinh(x/a) + c.

Completing the square and adjusting coefficients

x² + 4x + 13 = (x + 2)² + 9, so ∫ 1/(x² + 4x + 13) dx = (1/3) arctan((x + 2)/3) + c.

If x² has a coefficient, take it out: sqrt(4x² + 9) = 2 sqrt(x² + 9/4), so ∫ 1/sqrt(4x² + 9) dx = (1/2) arsinh(2x/3) + c.

Worked example

Evaluate ∫ from 2 to 4 of 1/sqrt(x² - 4) dx, giving an exact answer.

This is the arcosh form with a = 2
[arcosh(x/2)] from 2 to 4 = arcosh 2 - arcosh 1
arcosh 1 = 0 and arcosh 2 = ln(2 + sqrt(3))

Answer: ln(2 + sqrt(3))

Key idea: 1/(a² + x²) gives (1/a) arctan(x/a); 1/sqrt(a² - x²) gives arcsin(x/a); 1/sqrt(x² + a²) gives arsinh(x/a); 1/sqrt(x² - a²) gives arcosh(x/a). Complete the square first.

Check yourself

Evaluate ∫ from 0 to 2 of 1/(4 + x²) dx, to 3 decimal places.
Show answer
0.393. (1/2)[arctan(x/2)] from 0 to 2 = (1/2)(π/4) = π/8 = 0.393.
Evaluate ∫ from 0 to 3 of 1/sqrt(x² + 9) dx, to 3 decimal places.
Show answer
0.881. [arsinh(x/3)] from 0 to 3 = arsinh 1 = ln(1 + sqrt(2)) = 0.881.
Which is ∫ 1/sqrt(25 - x²) dx?
- arcsin(x/5) + c
- (1/5) arcsin(x/5) + c
- arsinh(x/5) + c
- (1/5) arctan(x/5) + c
Show answer
arcsin(x/5) + c. The form 1/sqrt(a² - x²) with a = 5 gives arcsin(x/a), with no 1/a factor.
Evaluate ∫ from -2 to 1 of 1/(x² + 4x + 13) dx, to 3 decimal places.
Show answer
0.262. (x + 2)² + 9, so (1/3)[arctan((x + 2)/3)] from -2 to 1 = (1/3)(π/4) = π/12 = 0.262.
Evaluate ∫ from 0 to 1 of 1/sqrt(4x² + 9) dx, to 3 decimal places.
Show answer
0.313. (1/2)[arsinh(2x/3)] from 0 to 1 = (1/2) arsinh(2/3) = 0.313.
Which substitution proves ∫ 1/sqrt(x² + a²) dx = arsinh(x/a) + c?
- x = a cosh u
- x = a sin u
- x = a tanh u
- x = a sinh u
Show answer
x = a sinh u. a² sinh² u + a² = a² cosh² u, so the square root becomes a cosh u, which cancels with dx = a cosh u du.
Evaluate ∫ from 0 to 1 of 1/sqrt(4 - x²) dx, to 3 decimal places.
Show answer
0.524. [arcsin(x/2)] from 0 to 1 = arcsin(1/2) = π/6 = 0.524.

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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.