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??

Question

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??

anonymous · Answer

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 · Answer

Indeed it did! Thank you very much!!

anonymous · Answer

You're welcome