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`
`(xdx)/(sqrt(7x-10-x^(2))^3)`can be evaluated by substituting for x as

Options

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

Answer

B

Explanation

`I=(dx)/(x+sqrt(x^(2)-x+1))`
Since here `c=1 gt 0, ` we can apply the second Euler substitution:
`sqrt(x^(2)-x+1)=tx-1`
` or (2t-1)x=(t^(2)-1)x^(2),x=(2t-1)/(t^(2)-1)`
Substituting into I, we get an integral of a rational fraction:
`int (dx)/(x+sqrt(x^(2)-x+1))=int(-2t^(2)+2t-2)/(t(t-1)(t+1)^(2))dt`
Now, `(-2t^(2)+2t-2)/(t(t-1)(t+1))=(A)/(t)+(B)/(t-1)+(D)/((t+1)^(2))+(E)/(t+1)`