How would you solve this? | Mathematics | Page 1

charred furnaceBOT Dec 11, 2023, 6:00 PM

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astral barn Dec 11, 2023, 6:08 PM

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how would we provide a simpler way if we don't know what way you used?

calm dragon Dec 11, 2023, 6:10 PM

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Id probably do a u sub

oblique sedge Dec 11, 2023, 6:13 PM

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I'm thinking integration by parts

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U sub gets you no where

calm dragon Dec 11, 2023, 6:13 PM

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No it will

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Yea I got it

oblique sedge Dec 11, 2023, 6:13 PM

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How?

calm dragon Dec 11, 2023, 6:14 PM

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If u is 3+2^x

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du is 2^x ln2 dx

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this also means that du is (u-3)ln2 dx

oblique sedge Dec 11, 2023, 6:14 PM

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Oh shit you right

calm dragon Dec 11, 2023, 6:15 PM

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Am I tripping

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And then you split the fraction

hard hamlet Dec 12, 2023, 12:57 AM

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calm dragon this also means that du is (u-3)ln2 dx

how?

calm dragon Dec 12, 2023, 12:58 AM

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cuz u is 3+2^x so u-3 is 2^x

hard hamlet Dec 12, 2023, 12:58 AM

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yes

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But how do you make the substitution if you need a 2^x?

calm dragon Dec 12, 2023, 1:00 AM

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you do a u substitution for 2^x +3 right

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so du is 2^x*ln(2) dx

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so du = (u-3)*ln(2) dx

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Did you get this?

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Hang on taking a pic

hard hamlet Dec 12, 2023, 1:05 AM

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yes but The problem is how you make the substitution.

calm dragon Dec 12, 2023, 1:06 AM

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And when you substitute 2^x+3 into integral you have 1/u

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and you also have du/ ((u-3)*ln(2))

hard hamlet Dec 12, 2023, 1:08 AM

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Yes but I need a 2^x in the denominator and I don't have it

calm dragon Dec 12, 2023, 1:08 AM

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How did you solve

hard hamlet Dec 12, 2023, 1:09 AM

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didnt

calm dragon Dec 12, 2023, 1:10 AM

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You don’t need 2^x in denominator because 2^x is it’s own derivative x ln(2)

hard hamlet Dec 12, 2023, 1:11 AM

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Ah, okay, I understand, thank you.

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.solved

charred furnaceBOT Dec 12, 2023, 1:11 AM

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Solved

Post marked as solved by @hard hamlet.

Use .unsolved if this was a mistake.

#How would you solve this?