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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 6

Lesson 12: Integrating functions using long division and completing the square

# Integration using long division

Here we do polynomial long division to make an integral more computable.

## Want to join the conversation?

• I was working to find the integral of 3/(2x+4) using u-sub. If I factor out the 2 from the denominator and set u = x+2 I get the following solution: 3/2 * ln(x+2). However, if I do not factor and let u = 2x+4, I get 3/2 * ln(2x +4). What am I missing? Thanks.
• Remember that a general antiderivative of a function (indefinite integral) always has a constant of integration c attached to it. Assuming the above integral was done correctly, there should be a c attached to both. Notice that the first solution is 3/2 * ln(x+2) +c and the second is 3/2 * ln(2x+4) + c. Now manipulate (3/2) ln(2x+4) + c to get (3/2) ln(2*(x+2) ) + c and you get (3/2) ln(2) + (3/2) ln(x+2) +c by log properties. Notice now that 3/2 * ln(2) can be absorbed into the constant of integration, because it is a real number. Thus, we get (3/2) ln(x+2) + c for both the first and second solutions.

Always remember to include the constant of integration.
• He did the long division as if this is something I should kinda know already is this covered in algebra or something?
• Hi, I need help solving a problem.
The problem is to find the Anti-Derivative of x^2 + 2x - 2 + 15/(2x+3)
The answer I came up with is : x^3/3 + x^2 - 2x + 15ln(2x+3) + C
However, the correct answer is coming up as :
:x^3/3 + x^2 - 2x + (15ln(2x+3))/2 + C
My answer is almost identical to the correct answer except for the last part.
I don't understand why the last expression is divided by 2.
Can someone explain this to me please?
Thank you.
• Notice that in the standard integral forms, you ALWAYS have a du. This is NOT just some notation that you can ignore, it is vital to getting the correct answer. The du is the derivative of whatever you call u -- it MUST be present or you cannot use that standard derivative form.
for the `∫ (1/u) du = ln(u) + C` form, you must have the derivative of the denominator you are calling u.
For`15/(2x+3)`, you declared `u` to be `2x+3`. That is fine, but you MUST have its derivative.
`u = 2x+3`
`du = 2 dx`
in other words `dx = ½ du`
So, to use this form I would need to do the following:
`∫ 15 dx / (2x+3)` ← though not necessary, factor out the 15 to avoid mistakes
`15∫ dx / (2x+3)` ← now replace what you have with the u and du.
`15∫ (½ du) / (2x+3)` ← we do this because dx = ½du
`15∫ (½ du) / u` ← no we need to make this EXACTLY match the standard form.
`(15/2) ∫ du/ u` ← I now have a standard form and can integrate
`(15/2) ln(u) + C` ← now back substitute
`(15/2) ln(2x+3) + C`
Of course, we wouldn't typically show all of these steps, i just included them so you could get the idea.

But the big thing to understand is this: In ALL of the standard integral forms there is a du. This is the derivative the whatever you call u. This MUST be present or you cannot use this form.

For beginners, using u-substitution to make absolutely sure you have an exact match is the wisest course. Though, once you become proficient, you can probably skip the substitution and go straight to the integration. Here is the same calculation the way that I do it (this involves putting a bracket around my du, making sure I have the derivative of what I am calling u, although I won't actually use any substitutions:
`∫ 15 dx / (2x+3)` ← The "u" is 2x+3, so the "du" is 2dx.
`∫ 15 (½) [2 dx] / (2x+3)` ← notice that I multiplied and divided by 2, so I haven't changed anything.
` (15/2) ∫ [2 dx] / (2x+3)` ← factored out the constants that are not in the standard integral form I am using.
`= (15/2) ln(2x+3) + C`
• why would he use u substitution the point where he got 2/x-1 it's just seems unnecessary
• If you're able to solve the antiderivative without u-sub, then great. This is meant to instruct those who just starting out with antiderivatives.
• In the video, "Integration using long division" the fraction 4/(2x-2) is simplified to 2*(1/(x-1)) to result in 2*ln(|x-1|). However, if you leave the fraction as 2*(2/(2x-2)) the result could be integrated as 2*ln(|2x-2|) which is a different function.

1) is simplification before integration mandatory in these cases?
2) If so, how do you avoid potentially changing a function when going to a derivative and back again?
• ln|2𝑥 − 2| = ln|(𝑥 − 1) ∙ 2| = ln|𝑥 − 1| + ln 2.

Thereby, ln|2𝑥 − 2| + 𝐶 is the same set of solutions as ln|𝑥 − 1| + 𝐶, and it doesn't really matter which representation we choose, it's just that (𝑥 − 1) is a little neater than (2𝑥 − 2), because it only involves one operation rather than two.
• https://imgur.com/a/qA1ScKo

In the above question for the integral of 1/(2x+6), if you factor out a 1/2 from the equation it becomes 1/2* integral of 1/(x+3) then doing u-sub you get 1/2*ln(x+3).

How do you know when to factor out something versus not factoring something out because the 2 answers are different?
• It's technically the same thing.
1 / 2 * ln(2x + 6) = 1 / 2 * [ln(2) + ln(x + 3)] = 1 / 2 * ln(x + 3) + C,
where C = 1 / 2 * ln 2. The two integrals have a constant difference, and are therefore technically equivalent.
• I don't know if someone asked this already, but what would you do if the denominator has a greater power than the numerator (like x^2 for the denominator and x for the numerator)
• it still might work to do long division, there are also cases it may not work with an equal or lower degree for the denominator as well, in which case you just want to try other ways of simplifying or rewriting the numerator and denominator.
• at why must -4/(-2x+2) be simplified? why is antiderivative not -4ln(|-2x+2|)? why am i getting different graphs for when simplified and when not?

thanks