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

Question

Paisley Johnson · Accepted Answer

#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#

William Abdalla · Answer

undefined To integrate $ \int x^4(17-4x^5)^5 \, dx $ using substitution, let $ u = 17 - 4x^5 $, then $ du = -20x^4 \, dx $.

Solving for $ dx $, we get $ dx = -\frac{1}{20x^4} \, du $.

Substituting these into the integral, we have:

$$ \int x^4(17-4x^5)^5 \, dx = \int -\frac{1}{20x^4} u^5 \, du $$

Now, integrate $ -\frac{1}{20x^4} u^5 $ with respect to $ u $.

$$ = -\frac{1}{20} \int \frac{1}{x^4} u^5 \, du $$

$$ = -\frac{1}{20} \int u^5x^{-4} \, du $$

$$ = -\frac{1}{20} \left( \frac{u^6}{6x^4} + C \right) $$

Substitute back for $ u $:

$$ = -\frac{1}{120} \left( \frac{(17-4x^5)^6}{6x^4} + C \right) $$