# Thread: One Integral; Two Solutions?

1. ## One Integral; Two Solutions?

This exercise comes from Calculus with Analytic Geometry, 5th edition, by Purcell and Varberg, section 8.1, problem 55:

$$\int\frac{x + 1}{9x^2 + 18x + 10}dx$$

The answer key gives $$\frac{1}{18}ln|9x^2 + 18x +10| + C$$ and I understand how they determined it: $$u = 9x^2 + 18x + 10, du = \frac{1}{18}(x + 1)dx, \frac{1}{18}\int\frac{du}{u}$$

However, before looking up the answer, I had taken another route and came up with a different answer, which I am not convinced is incorrect:

$$\int\frac{x + 1}{9x^2 + 18x + 10}dx, \int\frac{x + 1}{9(x + 1)^2 + 1}dx, \int\frac{dx}{9(x + 1) + 1}, \int\frac{dx}{9x + 10}, u = 9x + 10, du = 9dx, \frac{1}{9}\int\frac{du}{u}$$

This gives $$\frac{1}{9}ln|9x+10| +C$$

Noting the apparent difference between these answers, I wanted to see whether they differed by a constant, which would make them equally valid. I wrote this to determine the difference between these solutions across some domain values:

Code:
import numpy as np
def f(x):
return np.log(np.abs(9*x + 10))/9 - np.log(np.abs(9*x**2 + 18*x + 10))/18
x = np.arange(2000)
a = f(x)
Though the values in a seem to converge towards the end of the array, those near the beginning are different enough that I doubt the correctness of my answer. Does anyone see a problem with how I got my solution?

2. ## Re: One Integral; Two Solutions? Originally Posted by Zexuo This exercise comes from Calculus with Analytic Geometry, 5th edition, by Purcell and Varberg, section 8.1, problem 55:

$$\int\frac{x + 1}{9x^2 + 18x + 10}dx$$

The answer key gives $$\frac{1}{18}ln|9x^2 + 18x +10| + C$$ and I understand how they determined it: $$u = 9x^2 + 18x + 10, du = \frac{1}{18}(x + 1)dx, \frac{1}{18}\int\frac{du}{u}$$

However, before looking up the answer, I had taken another route and came up with a different answer, which I am not convinced is incorrect:

$$\int\frac{x + 1}{9x^2 + 18x + 10}dx, \int\frac{x + 1}{9(x + 1)^2 + 1}dx, \int\frac{dx}{9(x + 1) + 1}, \int\frac{dx}{9x + 10}, u = 9x + 10, du = 9dx, \frac{1}{9}\int\frac{du}{u}$$
Your second and third expressions above are not equal You can't cancel (x+1) … it's not a factor of the denominator.
This gives $$\frac{1}{9}ln|9x+10| +C$$

Noting the apparent difference between these answers, I wanted to see whether they differed by a constant, which would make them equally valid. I wrote this to determine the difference between these solutions across some domain values:

Code:
import numpy as np
def f(x):
return np.log(np.abs(9*x + 10))/9 - np.log(np.abs(9*x**2 + 18*x + 10))/18
x = np.arange(2000)
a = f(x)
Though the values in a seem to converge towards the end of the array, those near the beginning are different enough that I doubt the correctness of my answer. Does anyone see a problem with how I got my solution?
see in red

LOL thanks.

4. ## Re: One Integral; Two Solutions? Originally Posted by Zexuo This exercise comes from Calculus with Analytic Geometry, 5th edition, by Purcell and Varberg, section 8.1, problem 55:

$$\int\frac{x + 1}{9x^2 + 18x + 10}dx$$

The answer key gives $$\frac{1}{18}ln|9x^2 + 18x +10| + C$$ and I understand how they determined it:

$$u = 9x^2 + 18x + 10,$$

$\displaystyle du = \frac{1}{18}(x + 1)dx, \ \ \ \$ No. See ** below.

$$\frac{1}{18}\int\frac{du}{u}$$
As you noted elsewhere, $\displaystyle \ 9x^2 + 18x + 10 \ = \ 9(x + 1)^2 + 1. \ \$ So, it is always positive. Therefore, after initially writing the absolute value
bars as part of the formula, they can be dropped, because they are not needed. Parentheses can be put in their place.
Then the final answer may be written as:

$\displaystyle \frac{1}{18}ln(9x^2 + 18x +10) + C$

** $\displaystyle \ \ \ \ du \ = \ (18x + 18)dx \ = \ 18(x + 1)dx$

5. ## Re: One Integral; Two Solutions?

Note that, to verify the result of an integral, you can differentiate it and check that you get the integrand back. If you don't there is an error somewhere.