Asked by Mike
Calculate the following integral:
∫ (5x+3) / (x^2 +4x+7)
∫ (5x+3) / (x^2 +4x+7)
Answers
Answered by
oobleck
If u = x^2+4x+7, du = 2x+4
so, 5x+3 = 5/2 (2x+4) - 7
That gives you
∫ (5/2) du/u - 7∫1/(x^2+4x+7) dx
Now, x^2+4x+7 = (x+2)^2 + 3
So now let v = x+2, and dv = dx, and you have
∫ (5/2) du/u - 7∫1/(v^2+3) dv
Both of those are standard forms, so just crank them out.
Be sure to verify your answer at someplace like wolframalpha.com
so, 5x+3 = 5/2 (2x+4) - 7
That gives you
∫ (5/2) du/u - 7∫1/(x^2+4x+7) dx
Now, x^2+4x+7 = (x+2)^2 + 3
So now let v = x+2, and dv = dx, and you have
∫ (5/2) du/u - 7∫1/(v^2+3) dv
Both of those are standard forms, so just crank them out.
Be sure to verify your answer at someplace like wolframalpha.com
Answered by
Mike
Do you bring out - 7∫1/(x^2+4x+7) dx , because you have a common denominator? It would be like saying -7∫1/u right?
Answered by
oobleck
no common denominator
they are separate integrals, using separate substitutions. When trying to get du/d, there was a 7 left over.
they are separate integrals, using separate substitutions. When trying to get du/d, there was a 7 left over.
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