Two shapes worth spotting: the exponential pair, and the three surd forms

What to assume they know

• Integration by parts, from the previous topic, including the case that closes on itself
• The six quadratic-denominator formulas and completing the square, from the third topic of this module
• The integral of the secant, from the first topic of this module
• The standard formulae, especially the inverse sine row, from module m01
• Recognising a function and its own derivative sitting side by side
• Splitting an algebraic fraction whose numerator can be adjusted by adding and subtracting a constant

What they should be able to do

• State the exponential result and identify, in a given integrand, the function and its derivative that make it apply
• Derive that result from integration by parts, and say which term cancels against which
• Rearrange an integrand that is not obviously of that shape until it is
• State the three surd results and say which two are near-twins and how they differ
• Reproduce the chapter's derivation of the first surd result, including the step where the original integral reappears
• Name the three trigonometric substitutions the chapter offers as an alternative route
• Complete the square under a root and reduce the result to one of the three
• Compare the Summary's statement of the exponential result against the section's, and say which is correct

Where it usually goes wrong

• "The exponential result is a new technique." It is a pattern, produced by integration by parts once and then never re-derived. Presenting it as a fourth method alongside substitution, partial fractions and parts overstates it; the work is entirely in recognising the shape.
• "If the integrand is an exponential times something, the result applies." It applies when the something is a sum of a function and that function's own derivative. Exercise 7.6 Q21 is an exponential times a sine and is not this shape at all — it is the previous topic's self-closing case.
• "The shape either is or is not there." Example 22 (ii) and Exercise 7.6 Q17 and Q20 all have to be rearranged into it, by adding and subtracting inside a numerator. A student told to look for the shape and not told to manufacture it will abandon three of the seven items in this topic's slice.
• "The three surd results are three separate formulas." Two of them are the same formula with one sign changed, and the third is the only one that answers with an inverse function. Exercise 7.7 Q11's four options differ almost entirely within that near-twin, which is a fair warning about how the distinction gets examined.
• "The chapter derives all three surd results." It derives the first in full and asserts that the other two follow the same way, printing them without working. Say which is which; an explanation that implies all three are proved on the page is describing a different book.
• "Completing the square under a root is a different skill from completing it in a denominator." It is the same step. What differs is which list you land on afterwards — §7.4's six formulas from a denominator, §7.6.2's three results from a root over a whole integrand.
• "The trigonometric substitutions are a shortcut worth preferring." They are offered in one sentence as an alternative and no example uses them. If the explanation teaches them it should say the chapter names but does not demonstrate them.
• "The Summary can be trusted to restate the section." For the exponential result it does not. See section 10; this is the one place in the chapter where revising from the Summary produces a formula that cannot be used.

Questions to check understanding

• Identify the function and its derivative inside a given exponential integrand, and write down the integral
• Rearrange a numerator until the exponential shape appears — the form of Example 22 (ii) and of Exercise 7.6 Q17 and Q20
• State the three surd results and say which two are near-twins
• Derive the first surd result by parts, including the step that solves for the integral
• Complete the square under a root and reduce to the correct one of the three — the form of Examples 23 and 24 and of Exercise 7.7 Q3 to Q9
• Choose the correct integral from four options that differ only in a sign or a coefficient — the form of Exercise 7.7 Q11
• Distinguish an integrand of the exponential shape from an exponential times a sine, which is not

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.

• §7.6.1's derivation and result (Part II pp. 262–263). The heading itself names the shape: an exponential multiplying the sum of a function and that function's derivative. The chapter splits the integral in two, applies by parts to the first piece with the function as first factor and the exponential as second, and finds that the leftover term is exactly the second piece with the opposite sign. The answer is the exponential times the function, with no integral sign left. Read the result off the page image; it is set in bold there and it is what section 10 turns on.
• Example 22 (i) (Part II p. 263). An exponential multiplying an inverse tangent plus the reciprocal of one plus a square. The second summand is the derivative of the first, so the shape applies directly. Verified: the answer is the exponential times the inverse tangent.
• Example 22 (ii) (Part II p. 263). An exponential times a quadratic over a squared linear factor — and the shape is not visible until the numerator is rearranged. The chapter adds and subtracts inside the numerator until the fraction splits into two pieces, one of which is the derivative of the other. Verified by working added here: the function is the ratio of the variable less one to the variable plus one, its derivative is twice the reciprocal of the square of the variable plus one, the two pieces match, and the answer is the exponential times that ratio. This is section 4 and it is the only genuinely hard thing in the first half of the topic.
• Exercise 7.6, questions 16 to 20 (Part II pp. 263–264). Five items, all of them the §7.6.1 shape, and all of them needing the split spotted rather than performed: Q16 is a sine and a cosine and is immediate; Q17 and Q20 need Example 22 (ii)'s rearrangement; Q18 needs a Class XI half-angle rewrite before the split appears; Q19 is a reciprocal and its derivative and is the shortest in the set. Question 21, printed in the middle of this run, is not this shape — it is the previous topic's Example 21 with a doubled exponent — and is cited there.
• Exercise 7.6, questions 23 and 24 (Part II p. 264). Two multiple-choice items. Verified by working added here: Q23 is a plain substitution for a cube and has nothing to do with by parts at all, giving a third of the exponential of the cube, which is the first option; it is the odd item out in the whole exercise. Q24 is the §7.6.1 shape with the secant as the function and the secant times the tangent as its derivative, giving the exponential times the secant, which is the second option. The chapter prints no answers.
• §7.6.2's three results (Part II pp. 264–265), labelled (i), (ii) and (iii): the square root of a difference of the variable's square and a constant's; the same with a sum; and the square root of the constant's square less the variable's. Read off the page image, with the radicals and the modulus bars checked. The first two are near-twins — the same leading term, a logarithm in each, and only the sign under the root and the sign before the logarithm differing — while the third answers with an inverse sine instead. Compare them side by side; the two logarithmic ones are the pair students merge.
• The derivation of (i) (Part II pp. 264–265). The chapter takes the constant function one as the second factor, applies by parts, adds and subtracts the constant's square in the numerator that results, and finds the original integral back on the right, together with one of §7.4's six formulas. It then solves for the integral. Verified: the leftover integral is §7.4's fourth formula, and doubling the unknown and halving back gives exactly the printed result. The chapter says the other two go the same way and does not print them. Section 6 should run this one in full and section 7 should say plainly that the other two are asserted rather than derived here.
• The alternative route (Part II p. 265). One sentence offering three trigonometric substitutions instead — a secant for the first, a tangent for the second, a sine for the third. **This is the only place in the chapter that names the technique *trigonometric substitution***, and §7.4 used four of them without naming any.
• Example 23 (Part II p. 265). A general quadratic under a root, completed to a shifted square plus four, then matched to result (ii). Verified: the answer is half the shifted variable times the root, plus twice a logarithm of a modulus. The marginal citation on this page names the result correctly.
• Example 24 (Part II pp. 265–266). A general quadratic under a root with the variable's square negative, completed to four less a shifted square, then matched to result (iii). Verified: the answer is half the shifted variable times the root, plus twice an inverse sine of half the shifted variable. Together with Example 23 this is section 9, and the pair is the whole point: the same first step, and the sign under the root decides which of the three you land on.
• Exercise 7.7 (Part II p. 266), eleven items, of which the last two are multiple choice. Grouped by which of the three results the item reduces to — an added classification, not the chapter's: Q1, Q2 and Q5 land on the third, an inverse sine; Q3, Q4, Q7 and Q9 land on the second; Q6 and Q8 land on the first. Q9 needs a constant factored out from under the root before anything else, and Q8 has no constant term at all, so completing the square produces a shifted square with a subtracted constant.
• Exercise 7.7, questions 10 and 11 (Part II p. 266). Two multiple-choice items. Verified by working added here: Q10 is result (ii) with unit constant, giving half the variable times the root plus half a logarithm, which is the first option. Q11 completes to a shifted square less nine, so result (i) applies with the constant three; the answer is half the shifted variable times the root less nine halves of a logarithm, which is the fourth option. Note that the four options for Q11 differ only in the sign before the logarithm, in the coefficient on it and in one shift, which makes it a genuine test of the near-twin distinction rather than an arithmetic exercise. The chapter prints no answers.
• The Summary (Part II p. 290). It carries both halves of this topic, and one of them is printed wrongly. The three surd results appear as (i), (ii) and (iii) under a heading naming special types, and they match the section exactly. The exponential result appears as a bullet immediately above them, and its right-hand side keeps an integral sign and a differential where §7.6.1's own result has neither. See section 10 and the note below.

Figures to have open

• A three-row table of the surd results for section 5, a reduction table of Exercise 7.7 for section 11, and a two-column comparison of the section's exponential result against the Summary's for section 10. All three carry the chapter's own content, from Part II pp. 263–266 and Part II p. 290; the arrangements are added here. Build all three with the repo's DataTable component.
• A two-colour highlight of a function against its derivative inside one bracket, for sections 1 and 3, reused across three different integrands. Not in the book; the chapter prints the bracket and colours nothing.
• 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", §7.6.1 Integral of the type with an exponential, the derivation and the result, Part II pp. 262–263
• Example 22, two parts, Part II p. 263
• Exercise 7.6, questions 16 to 20 and 23 to 24, Part II pp. 263–264
• §7.6.2 Integrals of some more types, the three results, the derivation of the first, and the alternative trigonometric substitutions, Part II pp. 264–265
• Examples 23 and 24, Part II pp. 265–266
• Exercise 7.7, questions 1 to 11, Part II p. 266
• Summary, the exponential bullet and Some special types of integrals items (i) to (iii), Part II p. 290