Question
Find the indefinite integral of x^4/1-x^5dx.
so far, I have these steps:
∫ f'(x)dx = ∫ x^4/1-x^5dx
Let u=1-x^5
du= -5x^4
∫ u^-1*-1/5du (but I'm not sure if you can technically do -1/5du? this is where I'm getting lost. please help!!)
so far, I have these steps:
∫ f'(x)dx = ∫ x^4/1-x^5dx
Let u=1-x^5
du= -5x^4
∫ u^-1*-1/5du (but I'm not sure if you can technically do -1/5du? this is where I'm getting lost. please help!!)
Answers
Steve
everything is just perfect. -1/5 is just a number. Since
du = -5x^4 dx, the x^4 dx in the integrand becomes -1/5 du
Now just do the integral, and you have
-1/5 ln(u)+C
du = -5x^4 dx, the x^4 dx in the integrand becomes -1/5 du
Now just do the integral, and you have
-1/5 ln(u)+C