What is the integral of #(x^(-4) - sqrt(x^3) + 7)^2#?
1 Answer
I got:
#x^4/4 - 28/5x^(5//2) + 49x + 4/(3x^(3//2)) - 14/(3x^3) - 1/(7x^7) + C#
You should just expand this. I wouldn't integrate this as it is, because it can be made easier.
#= int (x^(-4) - sqrt(x^3) + 7)(x^(-4) - sqrt(x^3) + 7)dx#
#= int x^(-8) - x^(-4 + 3/2) + 7x^(-4) - x^(3/2 - 4) + x^3 - 7x^(3//2) + 7x^(-4) - 7x^(3//2) + 49dx# (remembering that exponents add when multiplying polynomial terms)
Combine terms and simplify exponents:
#= int x^(-8) - 2x^(-5//2) + 14x^(-4) + x^3 - 14x^(3//2) + 49dx#
I think we're good to integrate now using the reverse power rule.
#= x^(-7)/-7 - 2 cdot -2/3 x^(-3//2) + 14 cdot -1/3 x^(-3) + x^4/4 - 14 cdot 2/5 x^(5//2) + 49x#
We'll add in the constant at the end. For now, simplify terms.
#= -1/(7x^7) + 4/(3x^(3//2)) - 14/(3x^3) + x^4/4 - 28/5x^(5//2) + 49x#
And rearrange them in order of degree for convention.
#= color(blue)(x^4/4 - 28/5x^(5//2) + 49x + 4/(3x^(3//2)) - 14/(3x^3) - 1/(7x^7) + C)#
Hard to make it look any nicer! You can convince yourself that Wolfram Alpha gave the same thing.
