Chapter 7 · Integrals

Miscellaneous · Q16: $\int e^{3\log x}(x^4+1)^{-1}\,dx$

Miscellaneous ExerciseSimplify e^(3 log x) = x³, then substitute t = x⁴ + 1 · Miscellaneous Exercise

The question

Integrate the function:

e^{3\log x}\left(x^4+1\right)^{-1}

Exam answer

Let $I = \displaystyle\int e^{3\log x}(x^4 + 1)^{-1}\,dx$ ⁠.

Since $x > 0$ ⁠, $e^{3\log x} = e^{\log x^3} = x^3$ ⁠. So

I = \int\frac{x^3\,dx}{x^4 + 1}

Put $x^4 + 1 = t$ ⁠. Then $4x^3\,dx = dt$ ⁠.

I = \frac14\int\frac{dt}{t} = \frac14\log|t| + C

$\therefore\ I = \dfrac14\log(x^4 + 1) + C$ ⁠.

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

The one idea this problem is built on

First simplify: $3\log x = \log x^3$ ⁠, and $e^{\log x^3} = x^3$ ⁠. The integrand becomes $\frac{x^3}{x^4+1}$ ⁠, and the numerator $x^3$ is a quarter of the derivative $4x^3$ of the denominator. A fraction of the form $\frac{g'(x)}{g(x)}$ integrates to $\log|g(x)|$ ⁠.

Given, to find, and the plan

Given: $f(x) = e^{3\log x}(x^4+1)^{-1}$ ⁠. Because of $\log x$ ⁠, it is defined only for $x > 0$ ⁠.
To find: $\displaystyle\int e^{3\log x}(x^4+1)^{-1}\,dx$ ⁠.
Plan: (1) Simplify $e^{3\log x}$ ⁠. (2) Choose the substitution $t = x^4+1$ ⁠. (3) Integrate $\frac1t$ ⁠. (4) Return to $x$ ⁠. (5) Differentiate back and compare with the key.

Basics you need

1Laws of logarithms

$k\log x = \log x^k$ for $x > 0$ ⁠, and $e^{\log u} = u$ for $u > 0$ ⁠. Basics: logarithms and 1/x

2Substitution (Section 7.3.1)

$t = g(x)$ ⁠, $dt = g'(x)\,dx$ ⁠. In particular $\int\frac{g'(x)}{g(x)}\,dx = \log|g(x)| + C$ ⁠. Basics: substitution

3The logarithm integral

$\int\frac{dt}{t} = \log|t| + C$ (standard formula (xii), Section 7.2).

Solution

Simplify $e^{3\log x}$

For $x > 0$ ⁠:
$e^{3\log x} = e^{\log x^3} = x^3 ,\qquad f(x) = \frac{x^3}{x^4+1} .$
The power law of logarithms moves the 3 inside as an exponent, and then the exponential undoes the logarithm. Both laws need $x > 0$ ⁠, which is exactly where the given function is defined. The negative power $(x^4+1)^{-1}$ just means $\frac{1}{x^4+1}$ ⁠.

Basics: logarithms and 1/x
Choose the method: substitute $t = x^4 + 1$
$t = x^4 + 1 ,\qquad dt = 4x^3\,dx ,\qquad x^3\,dx = \frac{dt}{4} .$
The derivative of the denominator, $4x^3$ ⁠, is a constant multiple of the numerator. So the whole numerator together with $dx$ becomes $\frac{dt}{4}$ ⁠, and only $\frac1t$ is left to integrate. Partial fractions would be far longer here.

Section 7.3.1. Lesson: integration by substitution
Integrate in $t$
$\int\frac{x^3\,dx}{x^4+1} = \frac14\int\frac{dt}{t} = \frac14\log|t| + C .$
Standard formula (xii) with the constant $\frac14$ taken outside (Property (IV), Section 7.2.1).

Standard formula (xii), Section 7.2. Basics: logarithms and 1/x
Return to $x$
$\int e^{3\log x}(x^4+1)^{-1}\,dx = \frac14\log\left(x^4+1\right) + C .$
$t = x^4 + 1 \ge 1 > 0$ ⁠, so $|t| = t$ and the modulus can be dropped.
Check: differentiate the answer
$\begin{aligned}\frac{d}{dx}\left(\frac14\log(x^4+1)\right) &= \frac14\cdot\frac{4x^3}{x^4+1}\\ &= \frac{x^3}{x^4+1}\\ &= e^{3\log x}(x^4+1)^{-1}\ \checkmark\end{aligned}$
Chain rule: outer function $\log u$ ⁠, inner $u = x^4+1$ ⁠. The last equality is Step 1, which holds for $x > 0$ ⁠. The answer is therefore valid on $(0, \infty)$ ⁠, the whole domain of the given function.

Basics: integration undoes differentiation
Compare with the textbook's answer

The key gives $\frac14\log(x^4+1) + C$ ⁠, exactly our answer.

The formula $\frac14\log(x^4+1)$ would also be an antiderivative of $\frac{x^3}{x^4+1}$ for $x \le 0$ ⁠, but the function in the question, with its $\log x$ ⁠, does not exist there; so the answer is stated for $x > 0$ ⁠.

Result

For $x > 0$ ⁠:

\int e^{3\log x}(x^4+1)^{-1}\,dx = \frac14\log(x^4+1) + C

(Textbook answer: $\frac14\log(x^4+1) + C$ ⁠.)

Check it with numbers

At $x = 2$ ⁠: $e^{3\log 2} = e^{\log 8} = 8$ ⁠, so the integrand is $\frac{8}{17}$ ⁠, the same as $\frac{2^3}{2^4+1}$ ⁠.

Area from 1 to 2: $F(2) - F(1) = \frac14(\log 17 - \log 2)$ ⁠, that is $\frac14\log\frac{17}{2} \approx 0.5350165409$ ⁠. Numerical integration of $e^{3\log x}(x^4+1)^{-1}$ from 1 to 2 also gives $0.5350165409$ ⁠.

Common mistakes

Reading $e^{3\log x}$ as $3x$ ⁠. The 3 is a power: $e^{3\log x} = x^3$ ⁠.
Forgetting the $\frac14$ ⁠. $dt = 4x^3\,dx$ ⁠, so $x^3\,dx = \frac{dt}{4}$ ⁠.
Answering $\tan^{-1}(x^2)$ ⁠. That integrates $\frac{2x}{x^4+1}$ ⁠, a different function.

What to take away

$e^{k\log x} = x^k$ for $x > 0$ ⁠: simplify such factors before integrating.
If the numerator is a constant multiple of the derivative of the denominator, the integral is that constant times $\log|\text{denominator}|$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$k\log x = \log x^k$ ⁠; $e^{\log u} = u$	Step 1	Chapter 5, logarithmic functions (Basics)
Integration by substitution	Steps 2, 3	Section 7.3.1 (Lesson: substitution)
$\int\frac{dt}{t} = \log\|t\| + C$ ⁠; constants outside	Step 3	Standard formula (xii), Section 7.2; Property (IV), Section 7.2.1
Chain rule; $\frac{d}{dx}\log u = \frac{1}{u}$	Step 5	Chapter 5

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Miscellaneous · Q16: ∫e3log⁡x(x4+1)−1 dx\int e^{3\log x}(x^4+1)^{-1}\,dx∫e3logx(x4+1)−1dx

Basics you need

Solution

Simplify e3log⁡xe^{3\log x}e3logx

Choose the method: substitute t=x4+1t = x^4 + 1t=x4+1

Integrate in ttt

Return to xxx

Check: differentiate the answer

Compare with the textbook's answer

Every fact used, and where it comes from

Miscellaneous · Q16: $\int e^{3\log x}(x^4+1)^{-1}\,dx$

Simplify $e^{3\log x}$

Choose the method: substitute $t = x^4 + 1$

Integrate in $t$

Return to $x$