Exercise 7.2 · Q38 (MCQ): $\int\frac{10x^9+10^x\log_e 10}{x^{10}+10^x}\,dx$

Basics you need

Solution

1. The denominator is always positive

   For every real $x$⁠: $10^x > 0$ (an exponential is always positive) and $x^{10} \ge 0$ (an even power). So

   $$x^{10} + 10^x > 0 \quad\text{for all real } x.$$

   The integrand is defined for every real $x$⁠.

   This tells us the answer holds on the whole real line, and that $|x^{10} + 10^x| = x^{10} + 10^x$⁠, so option (D) can be written without a modulus.

2. Differentiate the denominator

   $$\frac{d}{dx}(x^{10} + 10^x) = 10x^9 + 10^x\log_e 10.$$

   This is exactly the numerator.

   Whenever a fraction has a complicated denominator, first try differentiating it; if the result matches the numerator (up to a constant), a logarithm is coming.
   Power rule $\frac{d}{dx}x^{10} = 10x^9$⁠; $\frac{d}{dx}10^x = 10^x\log_e 10$ (Example 28, Section 5.5, with $a = 10$⁠).

3. Substitute $t = x^{10} + 10^x$

   Put $t = x^{10} + 10^x$⁠, so $dt = (10x^9 + 10^x\log_e 10)\,dx$⁠. Then

   $$\int\frac{10x^9 + 10^x\log_e 10}{x^{10} + 10^x}\,dx = \int\frac{dt}{t} = \log|t| + C.$$

   Since $t > 0$ (Step 1),

   $$\int f(x)\,dx = \log(10^x + x^{10}) + C.$$
   The numerator times $dx$ is $dt$⁠, and the denominator is $t$ (Section 7.3.1). The modulus is dropped because $t$ is positive everywhere.
   Standard formula (xii), Section 7.2. Lesson: integration by substitution

4. Check: differentiate the answer back

   Chain rule, outer $\log u$⁠, inner $u = 10^x + x^{10}$⁠:

   $$\frac{d}{dx}\log(10^x + x^{10}) = \frac{10^x\log_e 10 + 10x^9}{10^x + x^{10}}.$$

   This is the integrand. ✓

   Property (I), Section 7.2.1: the derivative of an antiderivative is the integrand.
   Basics: integration undoes differentiation

5. Examine every option

   Differentiate each option and compare with the integrand. At $x = 1$ the integrand is $\frac{10 + 10\log_e 10}{1 + 10} = 3.002350\ldots$⁠.

   • (A) $\frac{d}{dx}(10^x - x^{10}) = 10^x\log_e 10 - 10x^9$⁠. At $x = 1$ it equals $23.02585 - 10 = 13.02585$⁠, not $3.00235$⁠. Wrong.
   • (B) $\frac{d}{dx}(10^x + x^{10}) = 10^x\log_e 10 + 10x^9$⁠: this is only the numerator. At $x = 1$ it is $33.02585$⁠, not $3.00235$⁠. It forgets the division by the denominator. Wrong.
   • (C) By the chain rule, $$\frac{d}{dx}(10^x - x^{10})^{-1} = -\frac{10^x\log_e 10 - 10x^9}{(10^x - x^{10})^2}.$$ At $x = 1$ this is $-\frac{13.02585}{81} = -0.16081$⁠, not $3.00235$⁠. It is also undefined wherever $10^x = x^{10}$ (for example at $x = 10$⁠), while the integrand is defined everywhere. Wrong.
   • (D) Its derivative is the integrand (Step 4). Correct.
   Differentiating each option is a complete test: an option is an antiderivative exactly when its derivative is the integrand. One sample point where the derivatives disagree is enough to reject an option; for (D) we checked the equality for every $x$⁠.

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