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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 • 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? • 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) • 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 • I am getting an extra +1/2 term compared with Sal's answer. I checked it with WolframAlpha and I get the same result as mine. Out of curiosity, I factored out from the denominator a (-2) and used u-sub (u = x - 1). I am trying to check for mistakes in my solution. Thanks for checking it out! • Can you factor out a (-2) from the denominator, changing the integral to (-1/2)*integral (x-5)/(x+1)? I get a different, but similar, answer: (-1/2)[x- 6 ln |x+1|] + C

Appreciate the clarification!  