- #1

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how:

integral: (2x+4)/(x^2+2x+3) with respect to x, of course

integral: (2x+4)/(x^2+2x+3) with respect to x, of course

- Thread starter dmission
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- #1

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how:

integral: (2x+4)/(x^2+2x+3) with respect to x, of course

integral: (2x+4)/(x^2+2x+3) with respect to x, of course

- #2

HallsofIvy

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- #3

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any help would be appreciated

- #4

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-spilt up the fraction

-take out the constants of integration

-do a u substitution for the 2x part of numerator

-use arctan for the 4 part of the numerator

you will notice the the part you completed the square looks almost like the arctan formula except you will need to change 2 into (sqrt2)

- #5

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what do you mean by take out the constants of integration?

- #6

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bump, really not sure what to do :S

- #7

rock.freak667

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[tex]= \int \frac{2x}{(x+1)^2+2}dx + \int \frac{4}{(x+1)^2+2}}dx[/tex]

Now just recall that:

[tex]\int \frac{1}{X^2+A^2} dx = \frac{1}{A}tan^{-1}(\frac{X}{A})+k[/tex]

- #8

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thanks for the reply.

I get how to take the second one, but what about the first? (2x/((x+1)^2+2))

I get how to take the second one, but what about the first? (2x/((x+1)^2+2))

- #9

rock.freak667

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Try a substitution of u=x+1thanks for the reply.

I get how to take the second one, but what about the first? (2x/((x+1)^2+2))

- #10

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yeah, but doesn't that still leave an X in the numerator?

- #11

rock.freak667

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u=x+1 => x=u-1yeah, but doesn't that still leave an X in the numerator?

- #12

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- #13

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2 * integral: (u-1)/(u^2+2), split up again,

and ultimately got:

ln((x+1)^2+2) - 2/sqrt(2)*arctan((x+1)/sqrt(2))), but apparently i'm wrong... help ?

- #14

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bump, anyone?

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