(1)Integral (3x-4)/ (x^2 - 4x +7). and (2) Integral (sinh 2x)/(Cosh 2x+1)^2
I've tried substitution, integrating by parts etc, but these two questions are still stumping me, help anyone?
I've tried substitution, integrating by parts etc, but these two questions are still stumping me, help anyone?
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1.) ∫(3x - 4)/(x^2 - 4x + 7) = ∫(2x + x - 4)/(x^2 - 4x + 7) dx
= ∫(2x - 4)/(x^2 - 4x + 7) dx + ∫x/(x^2 - 4x + 7) dx
= ln(x^2 - 4x + 7) + ∫(x - 2 + 2)/(x^2 - 4x + 7)
= ln(x^2 - 4x + 7) + 1/2*ln(x^2 - 4x + 7) + 2*∫dx/[(x - 2)^2 + 3]
= 3/2*ln(x^2 - 4x + 7) + 2/√3 * arctan((x - 2)/√3) + C
= ∫(2x - 4)/(x^2 - 4x + 7) dx + ∫x/(x^2 - 4x + 7) dx
= ln(x^2 - 4x + 7) + ∫(x - 2 + 2)/(x^2 - 4x + 7)
= ln(x^2 - 4x + 7) + 1/2*ln(x^2 - 4x + 7) + 2*∫dx/[(x - 2)^2 + 3]
= 3/2*ln(x^2 - 4x + 7) + 2/√3 * arctan((x - 2)/√3) + C
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this was a fun one.....
1. ((3^(3/2)) log (x^2 - 4x + 7) + 4 arctan((x-2)/(sqrt(3))))/(2^sqrt(3))
2. ((cosh2*sinh2(x) + sinh2)*log((cosh2(x) + 1) + sinh2)) / ((cosh2)^3(x) + (cosh2)^2)
I assume you know what cosh2 and sinh2 mean. they are imaginary..... you can read this article if you want to know more on substituion and so forth:
http://press.princeton.edu/books/maor/ch…
1. ((3^(3/2)) log (x^2 - 4x + 7) + 4 arctan((x-2)/(sqrt(3))))/(2^sqrt(3))
2. ((cosh2*sinh2(x) + sinh2)*log((cosh2(x) + 1) + sinh2)) / ((cosh2)^3(x) + (cosh2)^2)
I assume you know what cosh2 and sinh2 mean. they are imaginary..... you can read this article if you want to know more on substituion and so forth:
http://press.princeton.edu/books/maor/ch…
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Key it into Wolfram Alpha, it'll show you the steps.
Express the first in partial fractions.
Start with u=2x for the second.
Express the first in partial fractions.
Start with u=2x for the second.
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Wow...I thought I was having trig questions but then I saw yours...good luck!