$\displaystyle \int{(x^2 + 5x + 6)\sqrt{x + 1}}dx$ by the addition-subtraction method.

Moderator edit:Approved Challenge question.

Printable View

- Jan 14th 2011, 10:48 AMwonderboy1953Solve the following integral:
$\displaystyle \int{(x^2 + 5x + 6)\sqrt{x + 1}}dx$ by the addition-subtraction method.

**Moderator edit:**Approved Challenge question. - Jan 14th 2011, 11:35 AMdwsmith
Perhaps you can elaborate on this method since a google search on this technique provides nothing useful.

integration addition subtraction method - Google Search - Jan 14th 2011, 11:45 AMwonderboy1953
- Jan 14th 2011, 06:33 PMTheCoffeeMachine
I'm sort of confused here. What are you trying to achieve with this integral?

It's an absolute no-brainer. I'm just as confused about this alleged 'method'!

Letting $\displaystyle t = \sqrt{x+1}$ gives:

$\displaystyle \begin{aligned}& \int \left(x^2+5x+6\right)\sqrt{x+1}\;{dx} = \int {2(t^2+1)(t^2+2)t^2\;{dt} \\& = \int (2 t^6+6 t^4+4 t^2)\;{dt} = \frac{2}{7}t^7+\frac{6}{5}t^5+\frac{4}{3}t^3+k \\& = \frac{2}{7}\sqrt{(x+1)^7}+\frac{6}{5}\sqrt{(x+1)^5 }+\frac{4}{3}\sqrt{(x+1)^3}+k.\end{aligned}$ - Jan 14th 2011, 06:34 PMdwsmith
- Jan 15th 2011, 09:26 AMwonderboy1953Please recheck your algebra
A no-brainer for you and a few others perhaps (I could have given a much tougher problem). The A-S method is simpler/more efficient to apply to this type of problem, yet I've never seen it used in any math class nor text.

A simpler method would help avoid algebraic mistakes (btw please recheck your algebra as it needs fixing).

Since the A-S method hasn't been applied, my challenge still stands. You can try again TheCoffeeMachine or maybe someone else out there wants to take a whack at this problem (refer to the third post on this thread).

Since the A-S method hasn't been used to solve this problem, my challenge still stands and if you want to see details about this method, I'll just refer you again to the third post on this thread. - Jan 15th 2011, 10:11 AMTheCoffeeMachine
- Jan 15th 2011, 10:21 AMwonderboy1953
- Jan 15th 2011, 10:34 AMchiph588@
- Jan 15th 2011, 10:39 AMTheCoffeeMachine
Wrong! Let me show you in slower/detailed steps, my dear wonderboy.

We have $\displaystyle t = \sqrt{x+1} \Rightarrow \frac{dt}{dx} = \frac{1}{2\sqrt{x+1}} \Rightarrow dx = 2\sqrt{x+1}\;{dt} = 2t\;{dt}$. Similarly, we have

$\displaystyle t = \sqrt{x+1} \Rightarrow t^2 = x+1$ and $\displaystyle x^2+5x+6 = (x+2)(x+3) = (t^2+1)(t^2+2)$.

Thus $\displaystyle \int \left(x^2+5x+6\right)\sqrt{x+1}\;{dx} = \int {2(t^2+1)(t^2+2)t^2\;{dt}$. - Jan 15th 2011, 10:52 AMwonderboy1953Step-by-step explanation
TheCoffeeMachine substituted $\displaystyle t = \sqrt{x + 1}$

This means that $\displaystyle x = t^2 - 1$

When you do the substitution in the starting equation, you have:

$\displaystyle \int{[(t^2 - 1)^2 + 5(t^2 - 1) + 6]t\;{dt}}$ which leads to

$\displaystyle \int {(t^2+1)(t^2+2)t\;{dt}$ - Jan 15th 2011, 10:57 AMwonderboy1953Oops
I overlooked the differential so TheCoffeeMachine is right.

- Jan 18th 2011, 12:48 PMwonderboy1953Hint
For those who may want to see what I have in mind, you can factor the $\displaystyle (x^2 + 5x + 6)$ portion into $\displaystyle (x + 2)(x + 3)$. With the first factor, you can subtract and add 1 to change it into $\displaystyle [(x + 1) + 1]$ (do you see why?) What would you do with the second factor to go on to complete the problem using the addition-subtraction method?