Breaking a proper rational function into pieces with simple denominators

What to assume they know

• The standard formulae, especially the logarithm row and the inverse tangent row, from module m01
• Substitution, from the first topic of this module
• The six quadratic-denominator formulas, from the third topic of this module
• Polynomial long division, and the degree of a polynomial, from Class X
• Factorising a quadratic, and recognising one that does not factorise over the reals
• Solving two or three simultaneous linear equations in as many unknowns

What they should be able to do

• Decide whether a given rational function is proper or improper, using the chapter's own criterion
• Divide an improper rational function down to a polynomial plus a proper one
• Select the correct partial-fraction shape from the chapter's table, given a factorised denominator
• Determine the unknown numerators by matching coefficients across the two sides
• Explain why a decomposition is an identity rather than an equation, and what the distinction buys
• Handle a repeated linear factor, and say why one term is not enough for it
• Handle an unfactorisable quadratic factor, and say why its numerator has to be linear
• Recognise a denominator shape the chapter's table does not cover, and reach it by substituting first
• Say where the Summary's statement of the division rule differs from the section's, and which case falls through the gap

Where it usually goes wrong

• "Decompose first, divide later if needed." Division comes first, always, and Example 12 is the case. A decomposition attempted on an improper fraction produces a system with no solution and no diagnosis.
• "Improper means the numerator's degree is strictly larger." It means it is not strictly smaller, so the equal-degree case is improper too. Example 12 is exactly that case. The Summary states the strict version; section 10 exists because of it.
• "A squared factor needs one term with a squared denominator." It needs two terms, one with each power. With only the squared term the system is over-determined and has no solution; students read that as an arithmetic error.
• "A quadratic factor gets a constant numerator like everything else." It gets a linear one. Counting unknowns against equations makes this inevitable rather than arbitrary, and that counting is worth showing once.
• "Every denominator is covered by the table." Example 14's is not, and the chapter simply substitutes its way to a shape that is. A student told the table is exhaustive will get stuck on Exercise 7.5 Q18 and conclude the item is misprinted.
• "You may equate coefficients whenever two expressions are equal." Only when they are equal for every value — that is what the Remark on Part II p. 254 is for. Stating it after the first use, as the chapter does, teaches the move before the licence.
• "Substituting a root to find one unknown is a different method." It is the same identity used at a convenient point, and it is legitimate for exactly the reason the Remark gives. The chapter never does it; the explanation may, and should say the chapter does not.
• "The modulus bars can be dropped when the answer looks tidier." They can be dropped when the argument is provably positive, which is what Example 15 does and says. Anywhere else they stay.

Questions to check understanding

• Decide whether a stated rational function is proper or improper and act accordingly
• Divide an improper rational function and decompose the remainder — the form of Example 12 and of Exercise 7.5 Q6 and Q12
• Write down the correct decomposition shape for a given factorised denominator
• Determine the unknowns by comparing coefficients and integrate the result — the form of Exercise 7.5 Q1 to Q14
• Decompose across an unfactorisable quadratic factor and produce all three kinds of term — the form of Example 16 and of Exercise 7.5 Q11, Q13 and Q19
• Reach a denominator the table does not cover by substituting first — the form of Example 14 and of Exercise 7.5 Q18
• Choose the correct integral from four options — the form of Exercise 7.5 Q22 and Q23
• Justify comparing coefficients, by appeal to the decomposition being an identity

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• §7.5's opening (Part II pp. 252–253). It defines a rational function, calls it proper when the numerator's degree is the lower one and improper otherwise, says an improper one reduces by long division to a polynomial plus a proper one, and then restricts the whole section to denominators that factorise into linear and quadratic factors. That restriction is easy to miss and it is what makes the table finite. Read off the page image.
• Table 7.2 (Part II p. 253). Three columns — a serial number, the form of the rational function, and the form of its decomposition — and five rows, numbered in arabic. Read cell by cell off the page image. Row 1, two distinct linear factors with a linear numerator, decomposing into two constants over the two factors. Row 2, one linear factor squared, decomposing into a constant over the factor plus a constant over its square. Row 3, three distinct linear factors with a quadratic numerator, three constants. Row 4, one linear factor squared times a second distinct one, three constants over the factor, its square and the second factor. Row 5, a linear factor times a quadratic that cannot be broken further, giving a constant over the linear factor and a linear numerator over the quadratic. The row 5 condition is printed as a continuation line beneath the table body. Build the table with the repo's DataTable.
• Example 11 (Part II p. 254). The reciprocal of a product of two distinct linear factors, row 1. The chapter clears denominators, compares the coefficient of the variable and the constant term, and solves two equations. Verified: the two unknowns are one and minus one, and the answer is the logarithm of the modulus of the ratio of the two factors. See the note below about this page's marginal citation.
• The Remark on identities (Part II p. 254). It says the decomposition line is a statement true for every admissible value rather than for particular ones, and mentions that some authors mark the distinction with a separate symbol. This is the licence for comparing coefficients — you may only equate coefficients when the two sides agree everywhere — and the chapter states it once, in a Remark, after the technique has already been used. Section 5 should put it before the technique, not after.
• Example 12 (Part II pp. 254–255). A quadratic over a quadratic: equal degrees, therefore improper, so the chapter divides first and decomposes the remainder. Verified: the quotient is one, the remainder decomposes with unknowns minus five and ten, and the answer is the variable, less five times a logarithm, plus ten times a logarithm. This is the example that section 10 turns on — see the note below.
• Example 13 (Part II p. 255). A linear numerator over a squared linear factor times a distinct one, row 4. Three unknowns, three equations from the coefficients of the square, the first power and the constant. Verified by working added here: the three are eleven quarters, minus five halves and minus eleven quarters; the middle one integrates to a reciprocal rather than a logarithm, which is exactly what the squared factor is for. Checked independently by substituting the two roots into the cleared identity, which gives the second and third unknowns in one step each — a shortcut the chapter never uses.
• Example 14 (Part II p. 256). A square over a product of two quadratics, neither of which factorises. This shape is not one of Table 7.2's five rows. The chapter reaches it by substituting for the square, decomposing the result as a row 1 problem in the new letter, and then undoing the substitution before integrating. Verified: the two unknowns are minus a third and four thirds, and the answer is minus a third of an inverse tangent plus two thirds of another inverse tangent of half the variable. The chapter closes the example by saying the substitution was used only for the decomposition and not for the integration, which is the sentence section 8 is built on.
• Example 15 (Part II pp. 256–257). A trigonometric integrand that becomes rational under a substitution: setting the new letter equal to the sine turns the denominator into a perfect square of a linear factor, row 2. Verified: the two unknowns are three and four; the answer is three times a logarithm plus four over a linear expression, and the chapter then drops the modulus bars on the logarithm by observing that the argument never goes negative. That last step is worth attention — it is the only place in the chapter where a modulus is discharged by an argument about range rather than carried.
• Example 16 (Part II pp. 257–258). A quadratic over a linear factor times an unfactorisable quadratic, row 5. Verified by working added here: the three unknowns are three fifths, two fifths and one fifth, and the answer has three pieces — a logarithm of the linear factor, a logarithm of the quadratic, and an inverse tangent. Three kinds of term from one decomposition is the best summary of what row 5 does. See the note below about this page's marginal citation.
• Exercise 7.5 (Part II pp. 258–259), twenty-three items, of which the last two are multiple choice. Grouped by denominator — an added classification, not the chapter's: Q1 to Q5 and Q9 are distinct linear factors, rows 1 and 3; Q7, Q8, Q10 and Q14 carry a repeated linear factor, rows 2 and 4; Q11, Q13, Q19 and Q23 carry an unfactorisable quadratic, row 5; Q6 and Q12 are improper and need dividing first; Q15, Q18 and Q20 have denominators that must be factorised before any row applies; and Q16, Q17 and Q21 carry printed hints, two of them substitutions and one a multiply-through. Q18 is the shape of Example 14 — two quadratics, none of them in the table — and needs the same substitution.
• Exercise 7.5, questions 22 and 23 (Part II p. 259). Two multiple-choice items. Verified by working added here: Q22's integrand decomposes over two distinct linear factors with unknowns minus one and two, giving a logarithm of a squared factor over the other, which is the second option. Q23's integrand decomposes into a reciprocal of the variable less the variable over one plus its square, giving the logarithm of the modulus of the variable less half the logarithm of one plus its square, which is the first option. The chapter prints no answers.
• The Summary's partial-fractions block (Part II pp. 288–289). It restates the definition, gives the division rule, and reprints Table 7.2's five rows as items 1 to 5. Read off the page images. The five rows match the section's exactly. The division rule does not — see section 10 and the note below.

Figures to have open

• A five-row redraw of Table 7.2 for section 3, a shape grouping of Exercise 7.5 for section 11, and a two-column comparison of the two division rules for section 10. All three carry the chapter's own content, from Part II p. 253, Part II pp. 258–259 and Part II pp. 252 and 288; the groupings and the alignment are added here. Build all three with the repo's DataTable component.
• An unknowns-against-equations tally for section 7, showing why a quadratic factor forces a linear numerator. Not in the book; the chapter asserts the shape in the table and never counts.
• No figure is available from the chapter for any section of this topic. The whole chapter prints one figure, on Part II p. 267, and it belongs to module m03. Every picture listed here is an added construction over the chapter's own content.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 7 "Integrals", §7.5 Integration by Partial Fractions, the proper and improper definitions and the restriction on denominators, Part II pp. 252–253
• Table 7.2, five rows, Part II p. 253
• Examples 11 to 16, and the Remark on identities, Part II pp. 254–258
• Exercise 7.5, questions 1 to 23, Part II pp. 258–259
• Summary, Integration by partial fractions, Part II pp. 288–289