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. `log_(e)|t+1|`
B. `log_(e)|t+2|`
C. `(1)/(t+2)`
D. none of these

Answer

D

Explanation

Here, `a=1 gt 0.` Therefore, we make the substitution `sqrt(x^(2)+2x+2)=t-x.` Squaring both sides, we get
`2x+2tx=t^(2)-2 or x=(t^(2)-2)/(2(1+t)) or dx=(t^(2)+2t+2)/(2(1+t)^(2))dt`
`1+sqrt(x^(2)+2x+2)=1+t-(t^(2)-2)/(2(1+t))=(t^(2)+4t+4)/(2(1+t)).`
Substituting into the integral, we get
`I=int(2(1+t)(t^(2)+2t+2))/((t^(2)+4t+4)2(1+t)^(2))dt=int((t^(2)+2t+2)dt)/((1+t)(t+2)^(2))`
Now, let us expand the obtained proper rational fraction into partial fractions:
`(t^(2)+2t+2)/((t+1)(t+2)^(2))=(A)/(t+1)+(B)/(t+2)+(D)/((t+2)^(2)).`