Chapter 7 · Integrals

Exercise 7.6 · Q20: $\int\frac{(x-3)e^x}{(x-1)^3}\,dx$

Exercise 7.6Integral of the type eˣ[f + f′] · Section 7.6.1

The question

Integrate the functions in Exercises 1 to 22.

20.

\frac{(x-3)\,e^x}{(x-1)^3}

Exam answer

Let $I = \displaystyle\int\frac{(x - 3)\,e^x}{(x - 1)^3}\,dx$ ⁠.

\begin{aligned} I &= \int e^x\,\frac{(x - 1) - 2}{(x - 1)^3}\,dx \\ &= \int e^x\bigg[\frac{1}{(x - 1)^2} \\ &\qquad - \frac{2}{(x - 1)^3}\bigg]dx \end{aligned}

Let $f(x) = \frac{1}{(x - 1)^2}$ ⁠. Then ${f'(x) = -\frac{2}{(x - 1)^3}}$ ⁠.

Using $\displaystyle\int e^x\left[f(x) + f'(x)\right]dx$ $= e^xf(x) + C,$

$\therefore\ I = \dfrac{e^x}{(x - 1)^2} + C$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Write the top in terms of the bottom's base: $x - 3 = (x - 1) - 2$ ⁠. Then

\frac{x-3}{(x-1)^3} = \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3},

and $-\frac{2}{(x-1)^3}$ is the derivative of $\frac{1}{(x-1)^2}$ ⁠. So the integrand is $e^x[f + f']$ with $f = \frac{1}{(x-1)^2}$ ⁠, and Section 7.6.1 finishes it.

Given, to find, and the plan

Given: $\frac{(x-3)e^x}{(x-1)^3}$ ⁠, defined for $x \ne 1$ ⁠.
To find: $\int\frac{(x-3)e^x}{(x-1)^3}\,dx$ ⁠.
Plan: (1) Split the fraction. (2) Recognise $f = (x-1)^{-2}$ and apply the rule. (3) Differentiate back.

Basics you need

1The rule $\int e^x[f + f']\,dx = e^x f + C$

Section 7.6.1. Lesson: integration by parts

2Splitting a fraction

$\frac{a - b}{c} = \frac ac - \frac bc$ ⁠; a factor $x - 1$ may be cancelled because $x \ne 1$ ⁠. Basics: splitting fractions

3Derivative of $(x-1)^{-2}$

$\frac{d}{dx}(x-1)^{-2} = -2(x-1)^{-3}$ (power rule with the chain rule). Basics: powers of x

Solution

Split the fraction

Write $x - 3 = (x - 1) - 2$ ⁠. For $x \ne 1$ ⁠,
$\begin{aligned} \frac{x-3}{(x-1)^3} &= \frac{(x-1) - 2}{(x-1)^3} \\ &= \frac{1}{(x-1)^2} - \frac{2}{(x-1)^3}. \end{aligned}$
So the integral is $\int e^x\left[\frac{1}{(x-1)^2} - \frac{2}{(x-1)^3}\right]dx$ ⁠.

A cube $(x-1)^3$ in the bottom is what differentiating a square $(x-1)^{-2}$ produces. Writing the top as $(x - 1)$ plus a constant lets one factor cancel and exposes the two powers $-2$ and $-3$ ⁠.

Splitting a fraction. Basics: splitting fractions
Recognise $f + f'$ and apply the rule

Take $f(x) = \frac{1}{(x-1)^2} = (x-1)^{-2}$ ⁠. By the power rule and chain rule (inner function $x - 1$ ⁠, derivative 1),
$f'(x) = -2(x-1)^{-3} = -\frac{2}{(x-1)^3}.$
So the bracket is $f(x) + f'(x)$ ⁠, and by Section 7.6.1
$\int\frac{(x-3)e^x}{(x-1)^3}\,dx = \frac{e^x}{(x-1)^2} + C.$
$(x-1)^{-2}$ is differentiable on $(-\infty, 1)$ and on $(1, \infty)$ ⁠, so the answer holds on each of these intervals (the constant may differ between them).

$\int e^x[f + f']\,dx = e^x f + C$ (Section 7.6.1). Lesson: integration by parts
Check: differentiate the answer
$\begin{aligned} \frac{d}{dx}\left[\frac{e^x}{(x-1)^2}\right] &= \frac{e^x}{(x-1)^2} - \frac{2e^x}{(x-1)^3} \\ &= e^x\,\frac{(x-1) - 2}{(x-1)^3} \\ &= \frac{(x-3)\,e^x}{(x-1)^3}. \end{aligned}$
Product rule on $e^x\cdot(x-1)^{-2}$ ⁠, then put both terms over $(x-1)^3$ ⁠. The top is $x - 1 - 2 = x - 3$ ⁠: the integrand, for every $x \ne 1$ ⁠.

Product rule; chain rule (Chapter 5). Basics: integration undoes differentiation

Result

\int\frac{(x-3)e^x}{(x-1)^3}\,dx = \frac{e^x}{(x-1)^2} + C,

valid on any interval not containing $x = 1$ ⁠. (Textbook answer: $\frac{e^x}{(x-1)^2} + C$ ⁠, the same.)

Check it with numbers

On $[2, 3]$ ⁠, which does not contain 1:

\begin{aligned} F(3) - F(2) &= \frac{e^3}{4} - \frac{e^2}{1} \\ &= -2.3676718681\ldots \end{aligned}

Numerical integration of $\frac{(x-3)e^x}{(x-1)^3}$ over $[2, 3]$ gives $-2.3676718681\ldots$ ⁠, the same. The value is negative because $x - 3 \le 0$ on $[2, 3]$ while $e^x$ and $(x-1)^3$ are positive there.

Common mistakes

Writing $x - 3 = (x - 1) - 3$ ⁠. It is $(x - 1) - 2$ ⁠.
Derivative of $(x-1)^{-2}$ ⁠: it is $-2(x-1)^{-3}$ ⁠; the factor $-2$ must match the $-2$ in the split.
Ignoring $x = 1$ ⁠.

What to take away

With $e^x$ times $\frac{\text{linear}}{(x-a)^{k+1}}$ ⁠, write the top as a multiple of $(x - a)$ plus a constant; look for $f = \frac{c}{(x-a)^k}$ ⁠.
Questions 17, 19 and 20 are the same idea with different powers.

Every fact used, and where it comes from

Fact	Used in	Source
$\int e^x[f + f']\,dx = e^x f + C$	Step 2	Section 7.6.1 (Lesson: integration by parts)
Splitting a fraction; cancelling a nonzero factor	Step 1	Algebra
Power rule, chain rule, product rule	Steps 2, 3	Chapter 5, Sections 5.3 and 5.3.1

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Exercise 7.6 · Q20: ∫(x−3)ex(x−1)3 dx\int\frac{(x-3)e^x}{(x-1)^3}\,dx∫(x−1)3(x−3)ex​dx

Basics you need

Solution

Split the fraction

Recognise f+f′f + f'f+f′ and apply the rule

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q20: $\int\frac{(x-3)e^x}{(x-1)^3}\,dx$

Recognise $f + f'$ and apply the rule