anonymous
  • anonymous
Integral of 5x^3 dx / 1-4x^4 We have to solve this by substitution. I took 1-4x^4 as my value of u. Then I found du/dx =-8x^3 du = -8x^3 dx -8du = x^3 dx I don't know what to so next or how to substitute correctly. Am I even on the right track??
Mathematics
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anonymous
  • anonymous
Integral of 5x^3 dx / 1-4x^4 We have to solve this by substitution. I took 1-4x^4 as my value of u. Then I found du/dx =-8x^3 du = -8x^3 dx -8du = x^3 dx I don't know what to so next or how to substitute correctly. Am I even on the right track??
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
ok, you're right in making u=1-4x^4 which would make du=-8x^3dx Because your coefficient is 5 and not -8, you should multiply everything in the integral by -8/5 and divide by it outside the integral, which is the same as multiplying by -5/8 This will make you be taking the integral of (-5/8)(1/u)(du) Did that actually make sense?
anonymous
  • anonymous
Indeed it did! Thank you very much!!
anonymous
  • anonymous
You're welcome

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