Euler's substitution:
Integrals of the form `intR(x, sqrt(ax^(2)+bx+c))dx` are claculated with the aid of one of the following three Euler substitutions:
i. `sqrt(ax^(2)+bx+c)=t+-x sqrt(a)if a gt 0`
ii. `sqrt(ax^(2)+bx+c)=tx+-x sqrt(c)if c gt 0`
iii. `sqrt(ax^(2)+bx+c)=(x-a)t if ax^(2)+bx+c=a(x-a)(x-b)` i.e., if `alpha` is real root of `ax^(2)+bx+c=0`
`int(xdx)/((sqrt(7x-10-x^(2)))^(3))` can be evaluated by substituting for x as

Options

A. `x=(5+2t^(2))/(t^(2)+1)`
B. `x=(5-t^(2))/(t^(2)+2)`
C. `x=(2t^(2)-5)/(3t^(2)-1)`
D. none of these

Answer

A

Explanation

In this case, `a lt 0 and c lt 0`. Therefore, neither the first nor the second Euler substitution is applicable. But the quadratic `7x-10-x^(2)` has real roots ` alpha=2, beta=5.` Therefore, we use the third Euler substitution:
`sqrt(7x-10-x^(2))=sqrt((x-2)(5-x))=(x-2)t `
` or 5-x=(x-2)t^(2)`
` or x=(5+2t^(2))/(t^(2)+1)`