# Math Help - a question about Integration

1. ## a question about Integration

i can not understand how to integrate a power function like this one:
∫ (x&#178; - 1) / (x&#179; - 3x + 1 ) ^ 6 dx

i can not understand how to integrate a power function like this one:
∫ (x² - 1) / (x³ - 3x + 1 ) ^ 6 dx
Let y = x^3 - 3x +1.
Then dy = (3x^2 - 3)dx = 3(x^2 - 1)dx, so your integral becomes

(1/3)*Int(dy/y^6) = (1/3)*(1/-7)1/y^7 = -1/(21y^7) = -1/[21(x^3 - 3x + 1)].

-Dan

3. why did u multiply by 1/3
(1/3)*Int(dy/y^6)

why did u multiply by 1/3
(1/3)*Int(dy/y^6)
Because dy = 3(x^2 - 1)dx and what we have in the numerator of the integrand is only (x^2 - 1)dx.

-Dan

5. u multiplied by 1/3 to get rid of 3 ?in all of the solutions to these questions the author multiplied by a constant to get rid of a number, that what i can not understand why this multiplication is done and why we should get rid of the constant (3 for example)?

u multiplied by 1/3 to get rid of 3 ?in all of the solutions to these questions the author multiplied by a constant to get rid of a number, that what i can not understand why this multiplication is done and why we should get rid of the constant (3 for example)?
Specifically, what I am trying to do is replace
(x^2 - 1)dx
with an expression in y. Since we have that
dy = 3(x^2 - 1)dx

(x^2 - 1)dx = (1/3)dy.

So this gives us
(x^2 - 1)dx/(x^3 - 3x + 1)^6 = (1/3)dy/y^6

which we can sub in under the integration symbol.

-Dan