How do you integrate #intx(4x+5)^3# using substitution?

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Answer 1 · 2017-02-07T18:05:16

#=>1/80(4x + 5)^5 - 5/64(4x + 5)^4 + C#

Let #u = 4x + 5#. Then #du =4dx -> dx= (du)/4#. This also means that #x = (u - 5)/4#.

#=>int (u - 5)/4 u^3 (du)/4#

#=>1/16intu^4 - 5u^3du#

#=>1/16(1/5u^5 - 5/4u^4) + C#

#=>1/80u^5 - 5/64u^4 + C#

#=>1/80(4x + 5)^5 - 5/64(4x + 5)^4 + C#

Hopefully this helps!