How do you use substitution to integrate $int(x^4(17-4x^5)^5 dx)$ ?

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Answer 1 · 2018-07-22T16:59:27

$I=intx^4(17-4x^5)^5dx=int(17-4x^5)^5*x^4dx$
$I=-1/20int[17-4x^5]^5*d(17-4x^5)=-1/20(17-4x^5)^6/6+c$
$:.I=-1/120(17-4x^5)^6+c$

Here ,

$I=intx^4(17-4x^5)^5dx=int(17-4x^5)^5*x^4dx$

Subst. $17-4x^5=u=>-20x^4dx=du=>x^4dx=-1/20du$

So, $I=-1/20intu^5du$

$:,I=-1/20u^6/6+c$

$:.I=-1/120(17-4x^5)^6+c$

How do you use substitution to integrate int(x^4(17-4x^5)^5 dx)int(x^4(17-4x^5)^5 dx)?