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New Member
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Sep 24, 2010, 10:59 PM
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evaluate the following integrals
1. (integral sign) 2x+5(over) x(squared)+3x+6
2. (integral sign) x(cubed)+5x(squared)+4(over) x(x(squared)+3x+2)
****over is equivalent to the division sign
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New Member
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Sep 24, 2010, 11:07 PM
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can someone assist me with these integrals
the s like sign e(raised to the power of 2x)+x(raised to the power of 2) cos3x dx. And the other question is
the s sign 2x+1 divided by the square root of (x(raised to the power of 2)+3x+6)
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Uber Member
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Sep 25, 2010, 01:11 AM
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1.
This is close to the form:
but you have f'(x) here = 2x + 3. So, break the fraction into two parts first.
2. Try a long division first as the fraction isn't a simple fraction.
3.
Separate into two integrals and use by parts for the x^2cos(3x) one.
4.
Use integration by parts for this one.
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New Member
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Sep 25, 2010, 01:19 AM
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Comment on Unknown008's post
My best friend asked me to check on a solution for this question. I think he missed out a semester and is trying to teach himself from scratch. Would it be possible to give me the entire solution? Looking forward to your response
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Ultra Member
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Sep 25, 2010, 04:31 AM
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For the last one:
Break it up into:
For the first half, let
See how? It whittles down to
Now, continue with the second half.
For that part, complete the square on the quadratic in the radical and then make an appropriate substitution.
The first problem can be tackled by completing the square as well.
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New Member
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Sep 25, 2010, 05:43 AM
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Comment on galactus's post
For the second part, would it be equal to 1/4U+C?
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Ultra Member
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Sep 25, 2010, 06:05 AM
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No, it is a bit more involved than that.
We start with:
complete square:
Let
Then, we get:
Now, try a trig sub perhaps to finish:
Now, let
Make the subs and it becomes:
integrate and make the back substitutions. The solution involves ln
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New Member
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Sep 25, 2010, 06:28 AM
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Comment on galactus's post
OK cool. Thank you. Help greatly appreciated.
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New Member
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Sep 25, 2010, 11:27 AM
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can you confirm if these answers that we got are correct. We placed x=0 in the second question, and both sides equate to the same. But just wanted confirmation.
∫ (2x + 5)/(x² + 3x + 6) dx
= ∫ (2x + 3 + 2)/(x² + 3x + 6) dx
= ∫ (2x + 3)/(x² + 3x + 6) dx + 2 ∫ dx/(x² + 3x + 6)
= ∫ (2x + 3)/(x² + 3x + 6) dx + 2 ∫ dx/((x + 3/2)² + 15/4)
= ∫ (2x + 3)/(x² + 3x + 6) dx + 2 ∫ dx/((x + 3/2)² + √(15)/2)
= ln (x² + 3x + 6) + 2 (2/√15) arctan( 2(x + 3/2)/√15) + C
= ln (x² + 3x + 6) + (4/√15) arctan( 2(x + 3/2)/√15) + C
∫ (x³ + 5x² + 4)/[x(x² + 3x + 2) ] dx
= ∫ (x³ + 5x² + 4)/(x³ + 3x² + 2x) dx
= ∫ (x³ + 3x² + 2x + 2x² - 2x + 4)/(x³ + 3x² + 2x) dx
= ∫ (x³ + 3x² + 2x)/(x³ + 3x² + 2x) dx + ∫ (2x² - 2x + 4)/(x³ + 3x² + 2x) dx
Let
(2x² - 2x + 4)/(x³ + 3x² + 2x) = A/x + B/(x + 1) + C/(x + 2)
Then
2x² - 2x + 4 = A(x + 1)(x + 2) + Bx(x + 2) + Cx(x + 1)
2x² - 2x + 4 = x²(A + B + C) + x(3A + 2B + C) + (2A)
By comparing coefficients of different powers of x:
2A = 4 ==> A = 2
-2 = 3A + 2B + C ==> -2 = 6 + 2B + C ==> 2B + C = -8
2 = A + B + C ==> 2 = 2 + B + C ==> B + C = 0 ==> B = -C
==> -2C + C = -8 ==> C = 8 ==> B = 8
==> (2x² - 2x + 4)/(x³ + 3x² + 2x) = 2/x + 8/(x + 1) - 8/(x + 2)
continuation
= ∫ (x³ + 3x² + 2x)/(x³ + 3x² + 2x) dx + ∫ (2x² - 2x + 4)/(x³ + 3x² + 2x) dx
= ∫ 1 dx + ∫ (2/x + 8/(x + 1) - 8/(x + 2)) dx
= x + 2ln|x| + 8ln|x + 1| - 8ln|x + 2| + C
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Uber Member
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Sep 25, 2010, 12:03 PM
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Uber Member
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Sep 25, 2010, 12:13 PM
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2. B = -8, not 8
Then, your answer becomes:
x + 2ln|x| - 8ln|x + 1| - 8ln|x + 2| + C
Good! :)
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