How to Find the Integral of 1/(a² + x²)^(3/2)

The integral of 1/(a² + x²)^(3/2) with respect to x equals x / (a²·√(a² + x²)) + C, where a is a positive constant (a ≠ 0) and C is the constant of integration. This result comes from the trigonometric substitution x = a·tanθ, which converts the algebraic integrand into a simple cosine integral. If you search for the integral of 1 a 2 x 2 3 2, you are looking at a classic arctangent-family problem that appears constantly in calculus courses and physics applications such as electric fields along a charged rod or the gravitational field of a ring.

Recognize the Standard Form

The integrand 1/(a² + x²)^(3/2) has the signature shape of a Pythagorean expression: a sum of a constant squared and a variable squared, raised to a power. That pattern is exactly what trigonometric substitution is designed to eliminate, because the identity 1 + tan²θ = sec²θ turns a² + x² into a single squared term.

Here, a is a fixed positive constant and x is the variable of integration. Because the integrand is even in x, the antiderivative will be an odd function of x, which is a useful sanity check later.

The standard move is to set x = a·tanθ. This is the same family of substitution used for integrals like ∫ dx/(a² + x²), which produces an arctangent Paul’s Online Math Notes. The extra power of 3/2 changes the final answer from an inverse tangent to a rational function, but the opening step is identical.

Whenever you see √(a² + x²), x = a·tanθ is the canonical choice; for √(a² − x²) you use sine, and for √(x² − a²) you use secant Wolfram MathWorld. Recognizing which radical form you have is the entire decision-making step.

Apply Trigonometric Substitution

Set x = a·tanθ, so dx = a·sec²θ dθ. The key simplification is the base term:

a² + x² = a² + a²·tan²θ = a²(1 + tan²θ) = a²·sec²θ.

Raising this to the 3/2 power gives (a² + x²)^(3/2) = (a²·sec²θ)^(3/2) = a³·sec³θ, assuming a > 0 so that a³ is positive. This is the step most students mishandle: the exponent 3/2 distributes over both the a² and the sec²θ, producing a³ (not a² or a^(3/2)) and sec³θ.

Now substitute everything into the integral:

∫ dx/(a² + x²)^(3/2) = ∫ (a·sec²θ dθ) / (a³·sec³θ) = (1/a²) ∫ dθ/secθ = (1/a²) ∫ cosθ dθ.

The messy algebraic integrand has collapsed into a single cosine. That is the payoff of the substitution, and it is why this technique is taught as a template rather than a one-off trick.

Because tanθ ranges over all real numbers as θ runs through (−π/2, π/2), the substitution covers every real x. On that interval secθ > 0 and cosθ > 0, so no absolute-value subtleties arise.

Evaluate and Back-Substitute

The remaining integral is elementary: ∫ cosθ dθ = sinθ. Therefore

∫ dx/(a² + x²)^(3/2) = (1/a²)·sinθ + C.

The last task is expressing sinθ in terms of x. Draw a right triangle with angle θ, opposite side x and adjacent side a. Then the hypotenuse is √(a² + x²), so sinθ = opposite/hypotenuse = x/√(a² + x²). Equivalently, since tanθ = x/a, use sinθ = tanθ/√(1 + tan²θ) = (x/a)/√(1 + x²/a²) = x/√(a² + x²), after multiplying numerator and denominator by a.

Both routes give the same result, so the final answer is:

∫ dx/(a² + x²)^(3/2) = x / (a²·√(a² + x²)) + C.

This closed form is clean and compact, and it matches the tabulated result found in major references Wolfram MathWorld. Notice the dimensions: the numerator carries one power of x, the denominator carries x² times x, so the whole expression scales like 1/x² — consistent with the integrand’s behavior.

Worked Example and Check

Take a = 3 and integrate from x = 0 to x = 4. Then

∫₀⁴ dx/(9 + x²)^(3/2) = [ x / (9·√(9 + x²)) ]₀⁴ = 4 / (9·√(25)) − 0 = 4/45.

So the definite integral equals 4/45 ≈ 0.0889. Now verify the antiderivative by differentiation. Let F(x) = x / (a²·√(a² + x²)) with a = 3, so F(x) = x/(9√(9 + x²)). Apply the quotient rule:

F′(x) = [1·√(9 + x²) − x·(x/√(9 + x²))] / (9(9 + x²)).

Combine the numerator over √(9 + x²): [(9 + x²) − x²] / √(9 + x²) = 9/√(9 + x²). Dividing by 9(9 + x²) gives

F′(x) = 1 / (9 + x²)^(3/2),

which is exactly the original integrand. The derivative check passes, confirming both the algebra and the constant.

If a is negative, the substitution still works, but the simplification of (a²)^(3/2) should be written as |a|³. Since the integrand only depends on a², the cleanest convention is to state the result for a > 0 or to replace a² with |a|² in the final answer. Most textbooks assume a > 0 without comment. For practice with these substitution patterns, Khan Academy’s integration section offers a useful set of graded exercises Khan Academy.

Key Takeaways

• The antiderivative is ∫ dx/(a² + x²)^(3/2) = x / (a²·√(a² + x²)) + C for a > 0.
• Use the substitution x = a·tanθ, relying on the identity 1 + tan²θ = sec²θ.
• The term (a² + x²)^(3/2) simplifies to a³·sec³θ, which is where the 1/a² factor in the answer originates.
• After integrating ∫ cosθ dθ = sinθ, back-substitute using sinθ = x/√(a² + x²) from a right triangle.
• Always verify by differentiating; the derivative should return the original integrand exactly.
• The same substitution family handles √(a² − x²) (use sine) and √(x² − a²) (use secant).