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idiopathic
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#1
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#1
Is it possible to use integration by subsitution for

∫ x²/(4-x) dx
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Gaz031
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#2
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(Original post by endeavour)
Is it possible to use integration by subsitution for

∫ x²/(4-x) dx
Yes. Do you want me to tell you what substitution to make?
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idiopathic
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Report Thread starter 14 years ago
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Argh got it now thanks.
It was just me being stupid and failing to realise (a+b)/c = a/c + b/c
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Nima
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#4
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(Original post by endeavour)
Find ∫ x²/(4-x) dx
Let u = 4 - x --> du/dx = -1 --> dx/du = -1 --> dx = -1 du

Hence: ∫ x²/(4-x) dx
= ∫ [(4 - u)(4 - u)]/u . (-1) du
= - ∫ (u^2 - 8u + 16)/u du
= - ∫ u - 8 + 16(1/u) du
= - [(u^2)/2 - 8u + 16lnu] + k
= 8u - (u^2)/2 - 16lnu + k
= 8(4 - x) - [(4 - x)^2]/2 - 16ln(4 - x) + k
= 32 - 8x - [(4 - x)^2]/2 - 16ln(4 - x) + k
Let k + 32 = c:
--> ∫ x²/(4-x) dx = c - 8x - [(4 - x)^2]/2 - 16ln(4 - x)

Nima
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