Integration of Irrational Functions
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Direct answer
Roots inside an integrand are an instruction, not an obstacle: for fractional powers of x, set x = t^m where m is the LCM of the denominators of the exponents, which converts every radical into a whole power of t and turns the integral into a polynomial or rational integral. For integrands built on √((x - a)(b - x)) with x between a and b, the completion-of-square identity (x - a)(b - x) = ((b - a)/2)^2 - (x - (a + b)/2)^2 hands the answer directly to the arcsine: ∫dx/√((x - a)(b - x)) = arcsin((2x - a - b)/(b - a)) + C. Surds over linear terms respond to putting the whole surd equal to t; products like √(x - a) × √(x - b) complete squares on the difference instead, leading to logarithms.
What you must remember
- LCM substitution: for ∫R(x^(1/p), x^(1/q), ...) dx, put x = t^m with m = LCM(p, q, ...); dx = m t^(m-1) dt and every root becomes a power of t.
- Surd over linear: ∫R(x, (a + bx)^(1/n)) dx yields to t = (a + bx)^(1/n), i.e., x = (t^n - a)/b.
- Arcsine family: ∫dx/√((x - a)(b - x)) = arcsin((2x - a - b)/(b - a)) + C, valid for a < x < b.
- The companion result: ∫dx/√((x - a)(x - b)) = ln|x - (a + b)/2 + √((x - a)(x - b))| + C for x > b — the "b - x" versus "x - b" distinction decides arcsin versus log.
- Standard symmetry: ∫dx/((x - a)^1/2 (b - x)^1/2) is the same arcsine integral after writing the denominator as one square root.
- Reciprocal-surds type: ∫dx/(x(1 + x^n)^(1/2)) style integrals surrender to t^2 = 1 + x^n, a recurring NCERT Exemplar pattern.
- After substitution, simplify before integrating: factor the t-polynomial first; premature partial fractions on unfactored forms is where the algebra collapses.
One substitution from start to finish
Compute ∫dx/(√x + ∛x). The exponents 1/2 and 1/3 have LCM 6, so set x = t^6, dx = 6t^5 dt; √x = t^3 and ∛x = t^2. The integral becomes ∫6t^5/(t^3 + t^2) dt = ∫6t^3/(t + 1) dt after cancelling t^2. Divide: t^3/(t + 1) = t^2 - t + 1 - 1/(t + 1). Integrate term by term to get 6(t^3/3 - t^2/2 + t) - 6 ln|t + 1| + C, and return to x: 2x^(1/2) - 3x^(1/3) + 6x^(1/6) - 6 ln(1 + x^(1/6)) + C. Verify by differentiating the first term: 2 × (1/2) x^(-1/2) = x^(-1/2), and the t-machinery guarantees the remaining terms recombine into 1/(√x + ∛x). The workflow — spot exponents, form the LCM, substitute, cancel, divide — is exactly what a 3-minute Main question tests.
Where students slip
JEE Main samples this shelf with direct substitutions (√x, ∛x combos; (1 + x^4)^(1/4) over x) and the arcsine family with shifted intervals like [1, 3]. Advanced likes definite irrational integrals where the substitution must also flip the limits, and hybrid integrands (one root in the numerator, another in a denominator) demanding the LCM call. The characteristic errors: choosing m as one denominator instead of the LCM, leaving a residual root after substitution; forgetting dx = m t^(m-1) dt and integrating as though dx = dt; mixing up (x - a)(b - x) (arcsin territory, bounded interval) with (x - a)(x - b) (log territory, unbounded); and dropping the modulus in the logarithm. A final practical habit — after integrating in t, substitute back immediately and mark the + C; answers left in t earn no marks. This cluster belongs to the indefinite-integration strand of the syllabus, more prominent in Main than Advanced.
Frequently asked questions
Which substitution handles integrals containing x^(1/2) and x^(1/3) together?
x = t^6, the sixth power being the LCM of 2 and 3; every fractional power becomes an integer power of t and dx = 6t^5 dt.
What is ∫dx/√((x - a)(b - x)) for a < x < b?
arcsin((2x - a - b)/(b - a)) + C, from rewriting the product as ((b - a)/2)^2 - (x - (a + b)/2)^2.
Why do √((x - a)(b - x)) and √((x - a)(x - b)) lead to different answers?
The first is a capped square (arcsine, bounded interval); the second rewrites as a difference of squares (x - c)^2 - d^2, whose integral is a logarithm.
How is ∫dx/(x + √(x^2 + 1)) approached?
Rationalise the denominator first (multiply by x - √(x^2 + 1)), which separates it into a polynomial part and a standard ∫dx/√(x^2 + 1) logarithm.
What is the most common algebraic error after the t-substitution?
Forgetting the Jacobian factor — writing dx = dt instead of dx = m t^(m-1) dt — which silently destroys every coefficient in the answer.