Let us consider the integral of the foll...

Let us consider the integral of the following forms
`f(x_(1), sqrt(mx^(2)+nx+p))^(1//2)`
Case I If `m gt 0`, then put `sqrt(mx^(2)+nx+C)=u pm x sqrt(m)`
Case II If `p gt 0`, then put `sqrt(mx^(2)+nx+C)=u x pm sqrt(p)`
Case III If quadratic equation `mx^(2)+nx+p=0` has real roots `alpha` and `beta` there put `sqrt(mx^(2)+nx+p)=(x-alpha) u ` or `(x-beta)u`
`int ((x+sqrt(1+x^(2)))^(15))/(sqrt(1+x^(2))) dx` is equal to

`((x+sqrt(1+x^(2)))^(16))/(10)+c`

`(1)/(15(sqrt(1+x^(2)+x)))+c`

`(15)/(sqrt(1+x^(2))+x)+`c

`((sqrt(1+x^(2))+x))/(15)+c`

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The correct Answer is:

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Let us consider the integral of the following forms f(x_(1), sqrt(mx^(2)+nx+p))^(1//2) Case I If m gt 0 , then put sqrt(mx^(2)+nx+C)=u pm x sqrt(m) Case II If p gt 0 , then put sqrt(mx^(2)+nx+C)=u x pm sqrt(p) Case III If quadratic equation mx^(2)+nx+p=0 has real roots alpha and beta there put sqrt(mx^(2)+nx+p)=(x-alpha) u or (x-beta)u To evaluate int(dx)/((x-1)sqrt(-x^(2)+3x-2)) one of the most suitable substitution could be

`sqrt(-x^(2)+3x-2)=u`

`sqrt(-x^(2)+3x-2)=(ux sqrt(2))`

`sqrt(-x^(2)+3x-2)=u(1-x)`

`sqrt(-x^(2)+3x-2)=u(x+2)`

`-a`

`+- m^(2)`