# Solve the following integral:

• Jan 14th 2011, 10:48 AM
wonderboy1953
Solve the following integral:
$\int{(x^2 + 5x + 6)\sqrt{x + 1}}dx$ by the addition-subtraction method.

Moderator edit: Approved Challenge question.
• Jan 14th 2011, 11:35 AM
dwsmith
Quote:

Originally Posted by wonderboy1953
$\int{(x^2 + 5x + 6)\sqrt{x + 1}}dx$ by the addition-subtraction method.

Perhaps you can elaborate on this method since a google search on this technique provides nothing useful.

• Jan 14th 2011, 11:45 AM
wonderboy1953
Quote:

Originally Posted by dwsmith
Perhaps you can elaborate on this method since a google search on this technique provides nothing useful.

I'll refer you to this thread "The addition-subtraction puzzle (calculus)" in the puzzle section.
• Jan 14th 2011, 06:33 PM
TheCoffeeMachine
I'm sort of confused here. What are you trying to achieve with this integral?

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

\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 PM
dwsmith
Quote:

Originally Posted by TheCoffeeMachine
I'm sort of confused here. What are you trying to achieve with this integral?

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

\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}

I didn't even bother to look it up in the forum.
• Jan 15th 2011, 09:26 AM
wonderboy1953
Quote:

Originally Posted by TheCoffeeMachine
I'm sort of confused here. What are you trying to achieve with this integral?

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

\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}

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 AM
TheCoffeeMachine
Quote:

Originally Posted by wonderboy1953
A simpler method would help avoid algebraic mistakes (btw please recheck your algebra as it needs fixing).

Where? I've rechecked and I can't find anything that needs fixing! In fact, it's perfect! (Cool)
• Jan 15th 2011, 10:21 AM
wonderboy1953
Quote:

Originally Posted by TheCoffeeMachine
Where? I've rechecked and I can't find anything that needs fixing! In fact, it's perfect! (Cool)

Should be $\int {(t^2+1)(t^2+2)t\;{dt}$, not $\int {2(t^2+1)(t^2+2)t^2\;{dt}$
• Jan 15th 2011, 10:34 AM
chiph588@
Quote:

Originally Posted by wonderboy1953
Should be $\int {(t^2+1)(t^2+2)t\;{dt}$, not $\int {2(t^2+1)(t^2+2)t^2\;{dt}$

If you differentiate his answer you get the integrand.
• Jan 15th 2011, 10:39 AM
TheCoffeeMachine
Quote:

Originally Posted by wonderboy1953
Should be $\int {(t^2+1)(t^2+2)t\;{dt}$, not $\int {2(t^2+1)(t^2+2)t^2\;{dt}$

Wrong! Let me show you in slower/detailed steps, my dear wonderboy.

We have $t = \sqrt{x+1} \Rightarrow \frac{dt}{dx} = \frac{1}{2\sqrt{x+1}} \Rightarrow dx = 2\sqrt{x+1}\;{dt} = 2t\;{dt}$. Similarly, we have
$t = \sqrt{x+1} \Rightarrow t^2 = x+1$ and $x^2+5x+6 = (x+2)(x+3) = (t^2+1)(t^2+2)$.

Thus $\int \left(x^2+5x+6\right)\sqrt{x+1}\;{dx} = \int {2(t^2+1)(t^2+2)t^2\;{dt}$.
• Jan 15th 2011, 10:52 AM
wonderboy1953
Step-by-step explanation
Quote:

Originally Posted by chiph588@
If you differentiate his answer you get the integrand.

TheCoffeeMachine substituted $t = \sqrt{x + 1}$

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

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

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

$\int {(t^2+1)(t^2+2)t\;{dt}$
• Jan 15th 2011, 10:57 AM
wonderboy1953
Oops
I overlooked the differential so TheCoffeeMachine is right.
• Jan 18th 2011, 12:48 PM
wonderboy1953
Hint
Quote:

Originally Posted by wonderboy1953
$\int{(x^2 + 5x + 6)\sqrt{x + 1}}dx$ by the addition-subtraction method.

Moderator edit: Approved Challenge question.

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