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MATH
[1251]
Integration techniques

Rationalising substitutions

Changes of variable that turn an integrand with surds into a rational function.

A rationalising substitution changes the variable so that an integrand with surds or exponentials becomes rational, after which the partial-fraction methods apply. A typical choice is a root uu that clears all rational powers of one base, or an exponential substitution such as u=ex/2u=e^{x/2}. Not every algebraic integrand can be rationalised by a single substitution, so check the transformed integrand before committing to the method.

Rationalising the integrand

[METHOD]

Rationalise an integrand containing surds

For an integrand whose roots are rational powers of a common expression (or whose exponential exponents are rationally related so that u=ecxu=e^{cx} makes them rational powers), choose a new variable that makes the integrand rational. If the result is improper, divide; then decompose into partial fractions and integrate.

Replacing the radical is not enough. After rewriting both the integrand and dxdx, every occurrence of the new variable must have an integer exponent; only then do polynomial division and partial fractions apply. Forgetting to rewrite dxdx in terms of dudu gives the wrong powers of uu.

When several rational powers of xx appear, set x=uNx=u^N where NN is a common multiple of their denominators, so every power becomes an integer power of uu. Work on an interval where the original real powers and the chosen inverse are defined; when NN is even, x=uNx=u^N covers only x≥0x\ge0.

The trigonometric and hyperbolic table clears a2±x2\sqrt{a^2\pm x^2} and x2−a2\sqrt{x^2-a^2} by a named identity. Use a rationalising substitution for a surd that is not one of those three shapes.

  1. [1]:Integrate a rational function
  2. [2]:Substitutions for a2±x2\sqrt{a^2\pm x^2} and x2−a2\sqrt{x^2-a^2}
  3. [3]:∫dx1+x1/4\int\frac{dx}{1+x^{1/4}}
  4. [4]:∫dxe2x−1\int\frac{dx}{\sqrt{e^{2x}-1}}
  5. [5]:∫dx(1+e−x)3/2\int\frac{dx}{(1+e^{-x})^{3/2}}
[THEOREM]

Partial fractions decomposition

For a fixed denominator after cancellation and factorisation, every proper rational function ff has a unique partial-fraction expansion with terms of the form

A(x−a)k,Bx+C(x2+bx+c)k,\frac{A}{(x-a)^k},\qquad \frac{Bx+C}{(x^2+bx+c)^k},

where the quadratic x2+bx+cx^2+bx+c is irreducible over the real numbers (i.e. b2−4c<0b^2-4c<0).

  1. [1]:Proper rational function
  2. [2]:Partial fraction terms
  3. [3]:Integrate a rational function
  4. [4]:Distinct linear factors
  5. [5]:Repeated linear factors
  6. 3 more
[METHOD]

Integrate a rational function

To integrate a rational function on an interval containing no zero of its denominator:

  1. If the function is improper, use polynomial division to write it as a polynomial plus a proper rational function and integrate the polynomial.
  2. Decompose the proper part into partial fractions.
  3. Integrate each linear-factor term using u=ax+bu=ax+b: its first power gives Aaln⁡∣ax+b∣\frac{A}{a}\ln|ax+b|, while a power k>1k>1 gives A(ax+b)1−ka(1−k)\frac{A(ax+b)^{1-k}}{a(1-k)}.
  4. For q(x)=ax2+bx+cq(x)=ax^2+bx+c and a term ∫Bx+Cq(x)k dx\int\frac{Bx+C}{q(x)^k}\,dx, split the numerator into a multiple of q′(x)=2ax+bq'(x)=2ax+b and a constant remainder:
Bx+Cq(x)k=B2aq′(x)q(x)k+Dq(x)k,D=C−Bb2a.\frac{Bx+C}{q(x)^k} =\frac{B}{2a}\frac{q'(x)}{q(x)^k}+\frac{D}{q(x)^k}, \qquad D=C-\frac{Bb}{2a}.

Then integrate the derivative term directly and complete the square in the constant remainder.

  1. [1]:Partial fractions decomposition
  2. [2]:Proper rational function
  3. [3]:Partial fraction terms
  4. [4]:Linear numerator over a quadratic
  5. [5]:Rationalise an integrand containing surds

Fractional powers of xx

The simplest rationalising substitution is a root of xx, u=x1/Nu=x^{1/N}, which makes every fractional power an integer power of uu.

[EXAMPLE]

∫dx1+x1/4\int\frac{dx}{1+x^{1/4}}

For the real principal fourth root, take x≥0x\ge0 and work on an interval with x>0x>0 when differentiating the antiderivative.

Let I=∫dx1+x1/4I=\int\frac{dx}{1+x^{1/4}}. Set u=x1/4u=x^{1/4}, or u4=xu^4=x, so dx=4u3 dudx=4u^3\, du. Then

I=∫4u31+u du=4∫1+u3−11+u du.I=\int\frac{4u^3}{1+u}\, du=4\int\frac{1+u^3-1}{1+u}\, du.

The sum of cubes is 1+u3=(1+u)(1−u+u2)1+u^3=(1+u)(1-u+u^2), hence

I=4∫(1+u)(1−u+u2)−11+u du=4∫(1−u+u2−11+u) du=4u−2u2+43u3−4ln⁡∣1+u∣+C=4x1/4−2x+43x3/4−4ln⁡∣1+x1/4∣+C.\begin{aligned} I&=4\int\frac{(1+u)(1-u+u^2)-1}{1+u}\, du\\ &=4\int\left(1-u+u^2-\frac{1}{1+u}\right)\, du\\ &=4u-2u^2+\frac{4}{3}u^3-4\ln|1+u|+C\\ &=4x^{1/4}-2\sqrt{x}+\frac{4}{3}x^{3/4}-4\ln\bigl|1+x^{1/4}\bigr|+C. \end{aligned}

[exam]

  1. [1]:Rationalise an integrand containing surds

Exponential substitutions

For a surd built from an exponential, choose uu so that the expression under the root becomes a polynomial in uu.

[EXAMPLE]

∫dxe2x−1\int\frac{dx}{\sqrt{e^{2x}-1}}

For real values, the radicand requires x≥0x\ge0, but the denominator vanishes at x=0x=0. Thus the integrand and its antiderivative are considered on x>0x>0.

Let I=∫dxe2x−1I=\int\frac{dx}{\sqrt{e^{2x}-1}}. Set u=e2x−1>0u=\sqrt{e^{2x}-1}>0, so the denominator becomes uu and e2x=u2+1e^{2x}=u^2+1. Differentiating u2=e2x−1u^2=e^{2x}-1 gives 2u du=2e2x dx2u\,du=2e^{2x}\,dx. Hence

2u du=2(u2+1) dx  ⟹  dx=uu2+1 du.2u\, du=2(u^2+1)\, dx\implies dx=\frac{u}{u^2+1}\, du.I=∫1u⋅uu2+1 du=∫1u2+1 du=tan⁡−1u+C=tan⁡−1(e2x−1)+C.I=\int\frac{1}{u}\cdot\frac{u}{u^2+1}\, du=\int\frac{1}{u^2+1}\, du=\tan^{-1}u+C=\tan^{-1}\bigl(\sqrt{e^{2x}-1}\bigr)+C.

[exam]

  1. [1]:Rationalise an integrand containing surds

Setting u2u^2 equal to a positive expression clears every half-integer power of that expression at once, since each becomes an integer power of uu.

[EXAMPLE]

∫dx(1+e−x)3/2\int\frac{dx}{(1+e^{-x})^{3/2}}

Let I=∫dx(1+e−x)3/2I=\int\frac{dx}{(1+e^{-x})^{3/2}}. Set u2=1+e−xu^2=1+e^{-x}. Differentiating gives 2u du=−e−x dx2u\, du=-e^{-x}\, dx. Since e−x=u2−1e^{-x}=u^2-1, one has −e−x=1−u2-e^{-x}=1-u^2, so

2u du=(1−u2) dx  ⟹  dx=2u1−u2 du.2u\, du=(1-u^2)\, dx\implies dx=\frac{2u}{1-u^2}\, du.

Here u=1+e−x>1u=\sqrt{1+e^{-x}}>1 for every real xx. Hence (1+e−x)3/2=(u2)3/2=u3(1+e^{-x})^{3/2}=(u^2)^{3/2}=u^3 and 1−u2=(1−u)(1+u)1-u^2=(1-u)(1+u). Thus

I=∫2u(1−u)(1+u)u3 du=−2∫1(u−1)(u+1)u2 du,I=\int\frac{2u}{(1-u)(1+u)u^3}\, du=-2\int\frac{1}{(u-1)(u+1)u^2}\, du,

the minus sign coming from 1−u=−(u−1)1-u=-(u-1). Decompose

1(u−1)(u+1)u2=Au+Bu2+Cu−1+Du+1.\frac{1}{(u-1)(u+1)u^2}=\frac{A}{u}+\frac{B}{u^2}+\frac{C}{u-1}+\frac{D}{u+1}.

Clearing the denominator,

1=Au(u−1)(u+1)+B(u−1)(u+1)+Cu2(u+1)+Du2(u−1).1=Au(u-1)(u+1)+B(u-1)(u+1)+Cu^2(u+1)+Du^2(u-1).

Setting u=0u=0 gives B=−1B=-1. Setting u=1u=1 gives C=12C=\frac12. Setting u=−1u=-1 gives D=−12D=-\frac12. Setting u=2u=2 gives 1=6A+3B+12C+4D1=6A+3B+12C+4D, hence A=0A=0. Thus

I=2∫1u2 du−∫1u−1 du+∫1u+1 du=−2u−ln⁡∣u−1∣+ln⁡∣u+1∣+C=−21+e−x−ln⁡∣1+e−x−1∣+ln⁡∣1+e−x+1∣+C.\begin{aligned} I&=2\int\frac{1}{u^2}\, du-\int\frac{1}{u-1}\, du+\int\frac{1}{u+1}\, du\\ &=-\frac{2}{u}-\ln|u-1|+\ln|u+1|+C\\ &=-\frac{2}{\sqrt{1+e^{-x}}}-\ln\bigl|\sqrt{1+e^{-x}}-1\bigr|+\ln\bigl|\sqrt{1+e^{-x}}+1\bigr|+C. \end{aligned}

[exam]

  1. [1]:Rationalise an integrand containing surds

Rational functions

Proper fractions, partial fractions and the integrals they become.

Seen questions

Worked solutions on trigonometric integrals, average values, reduction formulae, substitutions and partial fractions.

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Rationalising the integrandRationalise an integrand containing surdsFractional powers of xx∫dx1+x1/4\int\frac{dx}{1+x^{1/4}}Exponential substitutions∫dxe2x−1\int\frac{dx}{\sqrt{e^{2x}-1}}∫dx(1+e−x)3/2\int\frac{dx}{(1+e^{-x})^{3/2}}