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Chapter 7 · Integrals

Exercise 7.6 · Q17: ∫xex(1+x)2 dx\int\frac{xe^x}{(1+x)^2}\,dx

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

Integrate the functions in Exercises 1 to 22.

17.

x ex(1+x)2\frac{x\,e^x}{(1+x)^2}
Exam answer

Let I=∫xex(1+x)2 dxI = \displaystyle\int\frac{xe^x}{(1 + x)^2}\,dx⁠.

I=∫ex (1+x)−1(1+x)2 dx=∫ex[11+x−1(1+x)2]dx\begin{aligned} I &= \int e^x\,\frac{(1 + x) - 1}{(1 + x)^2}\,dx \\ &= \int e^x\left[\frac{1}{1 + x} - \frac{1}{(1 + x)^2}\right]dx \end{aligned}

Let f(x)=11+xf(x) = \frac{1}{1 + x}⁠. Then f′(x)=−1(1+x)2f'(x) = -\frac{1}{(1 + x)^2}⁠.

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

∴ I=ex1+x+C\therefore\ I = \dfrac{e^x}{1 + x} + C⁠.

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

The one idea this problem is built on

Make the bracket look like f+f′f + f' by adding and subtracting 11 on top: x=(1+x)−1x = (1 + x) - 1⁠. Then x(1+x)2=11+x−1(1+x)2\frac{x}{(1+x)^2} = \frac{1}{1+x} - \frac{1}{(1+x)^2}⁠, and the second term is the derivative of the first. The rule ∫ex[f+f′] dx=exf+C\int e^x[f + f']\,dx = e^x f + C of Section 7.6.1 finishes it.

Given, to find, and the plan
Given
xex(1+x)2\frac{x e^x}{(1+x)^2}⁠, defined for x≠−1x \ne -1⁠.
To find
∫xex(1+x)2 dx\int\frac{x e^x}{(1+x)^2}\,dx⁠.
Plan
(1) Split the fraction. (2) Recognise f=11+xf = \frac{1}{1+x} and f′=−1(1+x)2f' = -\frac{1}{(1+x)^2}⁠; apply the rule. (3) Differentiate back.

Basics you need

1The rule ∫ex[f+f′] dx=exf+C\int e^x[f + f']\,dx = e^x f + C

Section 7.6.1; it holds on any interval where ff is differentiable, because ddx(exf)=ex(f+f′)\frac{d}{dx}(e^x f) = e^x(f + f')⁠. Lesson: integration by parts

2Splitting a fraction

a−bc=ac−bc\frac{a - b}{c} = \frac ac - \frac bc⁠, then cancel common factors (allowed when they are not zero). Basics: splitting fractions

3Derivative of 11+x\frac{1}{1+x}

ddx(1+x)−1=−(1+x)−2\frac{d}{dx}(1+x)^{-1} = -(1+x)^{-2}⁠, by the chain rule. Basics: powers of x

Solution

  1. Split the fraction

    Write the top as x=(1+x)−1x = (1 + x) - 1⁠. For x≠−1x \ne -1⁠,

    x(1+x)2=(1+x)−1(1+x)2=11+x−1(1+x)2.\begin{aligned} \frac{x}{(1+x)^2} &= \frac{(1+x) - 1}{(1+x)^2} \\ &= \frac{1}{1+x} - \frac{1}{(1+x)^2}. \end{aligned}

    So the integral is ∫ex[11+x−1(1+x)2]dx\int e^x\left[\frac{1}{1+x} - \frac{1}{(1+x)^2}\right]dx⁠.

    A bottom of (1+x)2(1+x)^2 is what you get by differentiating something with bottom 1+x1 + x⁠. Adding and subtracting 1 makes a copy of 1+x1 + x appear on top, which cancels one factor of the bottom (allowed because 1+x≠01 + x \ne 0⁠).
    Splitting a fraction over a common bottom. Basics: splitting fractions

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

    Take f(x)=11+xf(x) = \frac{1}{1+x}⁠. By the chain rule,

    f′(x)=−1(1+x)2.f'(x) = -\frac{1}{(1+x)^2}.

    So the bracket is f(x)+f′(x)f(x) + f'(x)⁠, and by Section 7.6.1

    ∫xex(1+x)2 dx=ex1+x+C.\int\frac{x e^x}{(1+x)^2}\,dx = \frac{e^x}{1+x} + C.
    The rule needs ff to be differentiable on the interval we work on. 11+x\frac{1}{1+x} is differentiable on (−∞,−1)(-\infty, -1) and on (−1,∞)(-1, \infty)⁠, so the answer holds on each of these intervals. (Strictly, the constant may be chosen separately on the two intervals, since the function is not defined at −1-1 to join them.)
    ∫ex[f+f′] dx=exf+C\int e^x[f + f']\,dx = e^x f + C (Section 7.6.1). Lesson: integration by parts

  3. Check: differentiate the answer

    ddx[ex1+x]=ex1+x−ex(1+x)2=ex (1+x)−1(1+x)2=xex(1+x)2.\begin{aligned} \frac{d}{dx}\left[\frac{e^x}{1+x}\right] &= \frac{e^x}{1+x} - \frac{e^x}{(1+x)^2} \\ &= e^x\,\frac{(1+x) - 1}{(1+x)^2} \\ &= \frac{x e^x}{(1+x)^2}. \end{aligned}
    Product rule on ex⋅(1+x)−1e^x\cdot(1+x)^{-1}⁠: the first term from differentiating exe^x⁠, the second from differentiating (1+x)−1(1+x)^{-1}⁠. Over the common bottom (1+x)2(1+x)^2⁠, the top is (1+x)−1=x(1 + x) - 1 = x⁠: the integrand, for every x≠−1x \ne -1⁠.
    Product rule; chain rule (Chapter 5). Basics: integration undoes differentiation
Result ∫xex(1+x)2 dx=ex1+x+C,\int\frac{x e^x}{(1+x)^2}\,dx = \frac{e^x}{1+x} + C,

valid on any interval not containing x=−1x = -1⁠. (Textbook answer: ex1+x+C\frac{e^x}{1+x} + C⁠, the same.)

Check it with numbers

On [0,1][0, 1]⁠, which does not contain −1-1⁠:

F(1)−F(0)=e2−11=e2−1=0.3591409142…\begin{aligned} F(1) - F(0) &= \frac{e}{2} - \frac{1}{1} \\ &= \frac e2 - 1 = 0.3591409142\ldots \end{aligned}

Numerical integration of xex(1+x)2\frac{x e^x}{(1+x)^2} over [0,1][0, 1] gives 0.3591409142…0.3591409142\ldots⁠, the same.

Common mistakes
  1. Integrating by parts with xexx e^x and (1+x)−2(1+x)^{-2} without a plan: it can be done, but the split in Step 1 makes the structure obvious.
  2. Sign of f′f'⁠: the derivative of 11+x\frac{1}{1+x} is −1(1+x)2-\frac{1}{(1+x)^2}⁠; the minus sign is exactly the minus sign in the split.
  3. Ignoring x=−1x = -1⁠. The answer is not valid on an interval that contains −1-1⁠.
What to take away
  1. For exe^x times a fraction with a squared bottom (x+a)2(x + a)^2⁠, split the top as (multiple of x+ax + a⁠) plus a constant; look for f=1x+af = \frac{1}{x+a}-type terms.
  2. Always differentiate your chosen ff to confirm that the other term really is f′f'⁠, sign included.

Every fact used, and where it comes from

FactUsed inSource
∫ex[f+f′] dx=exf+C\int e^x[f + f']\,dx = e^x f + CStep 2Section 7.6.1 (Lesson: integration by parts)
Splitting a fraction; cancelling a nonzero factorStep 1Algebra
Chain rule, product rule; ddxex=ex\frac{d}{dx}e^x = e^xSteps 2, 3Chapter 5, Sections 5.3 and 5.3.1
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