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November 27, 2014

November 27, 2014

Posted by **mathstudent** on Sunday, January 6, 2008 at 11:30pm.

- calculus -
**Reiny**, Monday, January 7, 2008 at 12:28ammessy, are you sure you typed correctly?

I was hoping the bottom would factor so I could use partial fractions.

the way it stands I ran it through an integration program to get

(√10/10)[tan^{-1}[√10(x+1)/5]

are you working at that sophisticated level of integration??

Is this highschool level ??

- calculus -
**Count Iblis**, Monday, January 7, 2008 at 9:19amYou can just write the denominator as:

2x^2 + 4x + 7 =

2 (x^2 + 2 x + 7/2) =

2 [(x + 1)^2 + 5/2]

Then you use the fact that the integral of dx/(x^2 + a^2) = 1/a arctan (x/a)

So, the integral is:

1/2 1/sqrt(5/2) arctan[(x+1)/sqrt(5/2)]=

Answer given by Reiny above.

- calculus -
**mathstudent**, Monday, January 7, 2008 at 11:34amThanks Reiny + Iblis!

This is from Wiley textbook "Calculus: Early Transcendentals Combined, 8th Edition", section 8.4. I think problem #41 (from memory).

I typed it right. The answer you two wrote matches the book, however I couldn't figure out how to do it. Makes sense now. Thanks

Reiny, thanks for writing that. I feel like an idiot getting stuck on these textbook problems sometimes.

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