Integral reducible to form: `int1/ sqrt(ax^2 + bx + c) dx`

Integral reducible to form: (1)/(sqrt(ax^(2)+bx+c))dx

Integral reducible to form: 1/(ax^(^^)2+bx+c)dx

int (ax^(2) + bx + c) dx

int sqrt(x)(ax^(2)+bx+c)dx

int (ax^(2)+bx+c)dx

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

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

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

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

int (2ax+b)/(sqrt(ax^2+bx+c))dx