How do you evaluate the integral using the substitution method? x/(4x^2 - 3)^0.4

Question

anonymous · Answer

u is the expression in the parenthesis of the denominator

anonymous · Answer

so du=8x dx

anonymous · Answer

then make an integral with u's in it

anonymous · Answer

$$\int\frac{x}{(4x^2-3)^{0.4}}dx$$ 
say u=(4x^2-3) then du/dx=8x or xdx=du/8
the integral becomes: $$\int\frac{1}{8(u)^{0.4}}du$$

anonymous · Answer

So you just put 8x in for u?

anonymous · Answer

No, @imron07 set it up perfectly... solve the u integral... then put the x's back in after taking the integral by substituting u=4x^2-3 where ever u is.

anonymous · Answer

you could see it as this$$\frac{1}{8}\int\limits_{}^{}u ^{-0.4}du$$use the power rule for integrals
after you are done, put the x's back in

anonymous · Answer

Oh ok thank you guys!

anonymous · Answer

No problem :)