Find #int(5x)/(7-2x)^3# dx using substitution u=3-5x^3. How is it solved?

1 Answer
May 23, 2017

#int (5x)/(7-2x)^3# #dx=35/(8(7-2x)^2)-5/(4(7-2x))+"c"#

Explanation:

Although the question asks to use the substitution #u=3-5x^3#, this approach is highly illogical and, as such, will be disregarded for a more convenient approach.

#int (5x)/(7-2x)^3# #dx#

Let #u=7-2x#

#du=-2dx#

#dx=-1/2du#

#x=1/2(7-u)#

#int (5x)/(7-2x)^3# #dx=int5/u^3(1/2(7-u))(-1/2)# #du#

#=5/4int(u-7)/u^3# #du=5/4int1/u^2-7/u^3# #du#

#=-5/(4u)+35/(8u^2)+"c"#

#=35/(8(7-2x)^2)-5/(4(7-2x))+"c"#