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 November 28th, 2011, 07:36 PM #1 Newbie   Joined: Nov 2011 Posts: 11 Thanks: 0 Solving Indefinite Integral? Hi guys, I was wondering how would you solve this equation: $\int 9x\sqrt{2x-1} dx$ I can already tell by u-subs it won't work and I don't remember learning how to solve problems like these in my calc homework or class. I don't want an answer but can you give me a clue on how to solve a problem like this? I'd like to be able to know the concept well (: Thank you!
 November 28th, 2011, 08:00 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Solving Indefinite Integral? We are given: $\int9x\sqrt{2x-1}\,dx=9\int x\sqrt{2x-1}\,dx$ Let: $u=2x-1\:\therefore\:du=2\,dx$ giving us: $\frac{9}{2}\int\frac{u+1}{2}\sqrt{u}\,du=\frac{9}{ 4}\int u^{\small{\frac{3}{2}}}+u^{\small{\frac{1}{2}}}\,d u$ Now the integration will be straightforward.
November 28th, 2011, 08:09 PM   #3
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Re: Solving Indefinite Integral?

Quote:
 Originally Posted by MarkFL We are given: $\int9x\sqrt{2x-1}\,dx=9\int x\sqrt{2x-1}\,dx$ Let: $u=2x-1\:\therefore\:du=2\,dx$ giving us: $\frac{9}{2}\int\frac{u+1}{2}\sqrt{u}\,du=\frac{9}{ 4}\int u^{\small{\frac{3}{2}}}+u^{\small{\frac{1}{2}}}\,d u$ Now the integration will be straightforward.

Where did you get the $\frac{u+1}{2}$ from, Sir?

 November 28th, 2011, 08:12 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Solving Indefinite Integral? With $u=2x-1$ solving for x gives: $u+1=2x$ $\frac{u+1}{2}=x$ Then replace x with this expression.
November 28th, 2011, 08:15 PM   #5
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Re: Solving Indefinite Integral?

Quote:
 Originally Posted by MarkFL With $u=2x-1$ solving for x gives: $u+1=2x$ $\frac{u+1}{2}=x$ Then replace x with this expression.

Ah, Thanks. My TA never told us we could do that -- I assumed he covered all the material on u-sub but I guess I was wrong. Thanks, Mark.
(:

 November 28th, 2011, 08:20 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Solving Indefinite Integral? Glad to help, and welcome to the forum!
 November 28th, 2011, 08:22 PM #7 Newbie   Joined: Nov 2011 Posts: 11 Thanks: 0 Re: Solving Indefinite Integral? Thanks ! I just hope I won't have a problem like this on my test this weekend, haha. Wouldn't have thought of it this way at all xD.
 November 28th, 2011, 08:30 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Solving Indefinite Integral? Here's a slightly different way to look at it... You have that x there and it needs to be replaced by an expression containing u, and you have said to let u = 2x - 1, so transform it like: $x=\frac{2x}{2}=\frac{(2x-1)+1}{2}=\frac{u+1}{2}$
 November 29th, 2011, 06:28 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,753 Thanks: 2136 Alternatively, use integration by parts (if familiar with that method).
November 29th, 2011, 08:47 AM   #10
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Re: Solving Indefinite Integral?

Hello, Jpyo!

Another approach . . .

Quote:
 $\int 9x\sqrt{2x-1}\,dx$

If the expression under the radical is linear,
[color=beige]. . [/color]we can let $u$ equal the entire radical.

$\text{Let }\,u \:=\:\sqrt{2x\,-\,1} \;\;\;\Rightarrow\;\;\;u^2 \:=\:2x\,-\,1\;\;\;\Rightarrow\;\;\;x \:=\:\frac{u^2\,+\,1}{2} \;\;\;\Rightarrow\;\;\;dx \:=\:u\,du$

$\text{Substitute: }\:\int 9\left(\frac{u^2\,+\,1}{2}\right)\,\cdot\,u\,\cdot \,(u\,du) \;\;=\;\;\frac{9}{2} \int(u^4\,+\,u^2)\,di$

[color=beige]. . . . . . . .[/color]$=\;\frac{9}{2}\,\left(\frac{u^5}{5}\,+\,\frac{u^3} {3}\right)\,+\,C \;\;=\;\;\frac{9}{2}\,\cdot\,\frac{u^3}{15}\,(3u^2 \,+\,5)\,+\,C$

$\text{Back-substitute: }\:\frac{3}{10}\,\!(\sqrt{2x\,+\,1})^3\,\big[3(2x\,-\,1)\,+\,5\big]\,+\,C \;=\;\frac{3}{10}\,\!(2x\,-\,1)^{\frac{3}{2}}\big[6x\,-\,3\,+\,5\big]\,+\,C$

[color=beige]. . . . . . . . . . . . [/color]$=\;\frac{3}{10}\,\!(2x\,-\,1)^{\frac{3}{2}}\,(6x\,+\,2)\,+\,C \;=\;\frac{3}{10}\,\!(2x\,-\,1)^{\frac{3}{2}}\,\cdot\,2(3x\,+\,1)\,+\,C$

[color=beige]. . . . . . . . . . . . [/color]$=\;\frac{3}{5}\,\!(2x\,-\,1)^{\frac{3}{2}}(3x\,+\,1)\,+\,C$

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