Integration of Irrational Functions and Euler Substitutions

Integrals containing roots are made rational by a substitution: for √(ax + b) use u = √(ax + b), for √(a² − x²) use x = a sin θ, and for √(ax² + bx + c) the Euler substitutions turn the integral into one of a rational function. In brief: ∫ dx/(x + √x) with u = √x becomes ∫ 2 du/(u + 1) = 2 ln(√x + 1) + C, and ∫ √(1 − x²) dx = (x√(1 − x²) + arcsin x)/2 + C. The first Euler substitution is √(ax² + bx + c) = t ± √a x for a > 0.

Roots in an integrand prevent the direct use of the standard formulas. Three families of techniques remove them: linear substitutions for roots of linear expressions, trigonometric (and hyperbolic) substitutions for roots of quadratics of special form and the Euler substitutions for the general quadratic under the root. This guide explains each method with worked examples, shows the standard integrals that result and gives a table of substitutions.

How do you integrate roots of linear expressions?

For an integrand with ⁿ√(ax + b) (possibly several roots with exponents p/q), substitute u = ^N√(ax + b), where N is the least common multiple of the root indices. Then x = (u^N − b)/a is a polynomial in u, and the integral becomes rational.

Example 1

∫ dx/(x + √x): u = √x, x = u², dx = 2u du. ∫ 2u du/(u² + u) = ∫ 2 du/(u + 1) = 2 ln(u + 1) + C = 2 ln(√x + 1) + C.

Example 2: two different roots

∫ dx/(√x + ³√x): the least common multiple of 2 and 3 is 6. u = ⁶√x, x = u⁶, dx = 6u⁵ du, √x = u³, ³√x = u². The integral is ∫ 6u⁵ du/(u³ + u²) = ∫ 6u³ du/(u + 1). Divide: u³/(u + 1) = u² − u + 1 − 1/(u + 1). So 6(u³/3 − u²/2 + u − ln|u + 1|) + C = 2u³ − 3u² + 6u − 6 ln|u + 1| + C with u = ⁶√x.

Example 3: with a linear fraction

For roots of linear fractions √((ax + b)/(cx + d)), the substitution u = √((ax + b)/(cx + d)) makes the integrand rational. Take ∫ √((1 + x)/(1 − x)) dx with u = √((1 + x)/(1 − x)). Then x = (u² − 1)/(u² + 1) and dx = 4u du/(u² + 1)². The integral becomes ∫ 4u² du/(u² + 1)².

How do trigonometric substitutions work?

For integrals with one of the roots below, the substitution removes the root by a Pythagorean identity.

• √(a² − x²): x = a sin θ, √(…) = a cos θ, dx = a cos θ dθ.
• √(a² + x²): x = a tan θ, √(…) = a sec θ, dx = a sec²θ dθ.
• √(x² − a²): x = a sec θ, √(…) = a tan θ, dx = a sec θ tan θ dθ.

A general quadratic is first reduced to one of these forms by completing the square.

Example 4: √(1 − x²)

∫ √(1 − x²) dx: x = sin θ, dx = cos θ dθ. ∫ cos²θ dθ = θ/2 + sin 2θ/4 + C = (θ + sin θ cos θ)/2 + C = (arcsin x + x√(1 − x²))/2 + C. This is the area under a circle: ∫₀¹ √(1 − x²) dx = π/4, the area of a quarter of the unit disc.

Example 5: dx/√(x² + 4)

x = 2 tan θ, dx = 2 sec²θ dθ, √(x² + 4) = 2 sec θ. ∫ sec θ dθ = ln|sec θ + tan θ| + C = ln|√(x² + 4)/2 + x/2| + C = ln|x + √(x² + 4)| + C′. In general ∫ dx/√(x² ± a²) = ln|x + √(x² ± a²)| + C.

Example 6: x²/√(4 − x²)

x = 2 sin θ: ∫ 4 sin²θ · 2 cos θ dθ/(2 cos θ) = ∫ 4 sin²θ dθ = 2θ − sin 2θ + C = 2 arcsin(x/2) − (x/2)√(4 − x²) + C.

Example 7: completing the square

∫ dx/√(3 + 2x − x²) = ∫ dx/√(4 − (x − 1)²) = arcsin((x − 1)/2) + C.

Example 8: √(x² − 9)/x

x = 3 sec θ: ∫ (3 tan θ/(3 sec θ)) 3 sec θ tan θ dθ = 3 ∫ tan²θ dθ = 3(tan θ − θ) + C = √(x² − 9) − 3 arccos(3/x) + C (for x > 3).

What are the Euler substitutions?

For ∫ R(x, √(ax² + bx + c)) dx, with R a rational function, Euler found three substitutions that give a rational integrand in t:

1. First (a > 0): √(ax² + bx + c) = t − √a x. Squaring: ax² + bx + c = t² − 2√a x t + ax², so x = (t² − c)/(b + 2√a t), which is rational in t.
2. Second (c > 0): √(ax² + bx + c) = xt ± √c. Squaring and cancelling c gives x = (±2√c t − b)/(a − t²).
3. Third (the quadratic has real roots x₁, x₂): √(a(x − x₁)(x − x₂)) = t(x − x₁). Then a(x − x₂) = t²(x − x₁), which is solved for x as a rational function of t.

Example 9: first substitution

∫ dx/√(x² + 1): a = 1, so √(x² + 1) = t − x. Squaring: x² + 1 = t² − 2tx + x², so x = (t² − 1)/(2t) and dx = (t² + 1)/(2t²) dt, with √(x² + 1) = t − x = (t² + 1)/(2t). The integrand becomes dt/t. The integral is ln|t| + C = ln(x + √(x² + 1)) + C, the inverse hyperbolic sine, consistent with Example 5.

Example 10: third substitution

∫ dx/√((x − 1)(x − 2)) has two real roots, so the third substitution applies: √((x − 1)(x − 2)) = t(x − 1). Squaring and dividing by x − 1: x − 2 = t²(x − 1), so x = (t² − 2)/(t² − 1), which is rational in t. For this integral, completing the square is shorter: x² − 3x + 2 = (x − 3/2)² − 1/4, so by the standard formula ∫ dx/√((x − 3/2)² − 1/4) = ln|x − 3/2 + √(x² − 3x + 2)| + C.

Example 11: an application

The arc length of the parabola y = x² from 0 to 1 is ∫₀¹ √(1 + 4x²) dx. With 2x = tan θ: (1/2) ∫ sec³θ dθ = (1/4)[sec θ tan θ + ln|sec θ + tan θ|], evaluated from θ = 0 to arctan 2: (1/4)(√5 · 2 + ln(√5 + 2)) = (1/4)(4.472 + 1.444) = 1.479. The arc length is about 1.479.

What are the standard integrals with roots?

• ∫ dx/√(a² − x²) = arcsin(x/a) + C.
• ∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) arcsin(x/a) + C.
• ∫ dx/√(x² + k) = ln|x + √(x² + k)| + C.
• ∫ √(x² + k) dx = (x/2)√(x² + k) + (k/2) ln|x + √(x² + k)| + C.

Table of substitutions

Where do the methods come from?

The substitutions bear the name of Leonhard Euler, who published them in his Institutionum calculi integralis (1768–1770). The trigonometric substitutions follow from the geometry of the circle and the hyperbola and were in use by Newton and Leibniz. The question of which integrals with roots can be expressed with elementary functions was answered in the nineteenth century: integrals with the square root of a quadratic always can, while those with the square root of a cubic or quartic (elliptic integrals) generally cannot.

Where are they used?

Roots under integrals appear in arc length (√(1 + f′²)), in the area of circles and ellipses, in the period of a pendulum (an elliptic integral) and in relativity, where energies contain √(1 − v²/c²). The trigonometric substitutions are standard in physics for the potential of a charged rod, which gives ∫ dx/(x² + a²)^3/2 = x/(a²√(x² + a²)) + C.

Which mistakes are common?

Forgetting to convert dx

With x = a sin θ, dx = a cos θ dθ. Missing this factor changes the integral.

Wrong substitution for the sign

√(x² − a²) needs sec θ, not sin θ. Match the form to the identity: 1 − sin² = cos², 1 + tan² = sec², sec² − 1 = tan².

Writing √(cos²θ) = cos θ without a check

√(cos²θ) = |cos θ|. The substitution x = a sin θ with θ in [−π/2, π/2] makes cos θ ≥ 0, so it is valid.

Back-substituting incorrectly

Use a right triangle: for x = 2 sin θ, sin θ = x/2, cos θ = √(4 − x²)/2.