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

Evaluate int_(1)^(4) (ax^(2)+bx+c)dx .

Solve: 1. int_0^3(ax^2+bx+c)dx 2. int_-1^1e^xdx 3. int_(-pi//2)^(pi//2) cosx dx 4. int_0^10sec^2(3x+6)dx

If int_-2^-1 (ax^2-5)dx=0 and 5 + int_1^2 (bx + c) dx = 0 , then

Let a, b, c be non zero numbers such that int_(0)^(3)sqrt(x^(2)+x+1)(ax^(2)+bx+c)dx=int_(0)^(5)sqrt(1+x^(2)+x)(ax^(2)+bx+c)dx . Then the quadratic equation ax^(2)+bx+c=0 has

int(dx)/(ax^(2)+bx+c)=k_(1)tan^(-1)(x+A)/(B)+C if

Differentiate sin^(n)(ax^(2)+bx+c)

int (ax +b)^(2)dx

Differentiate w.r.t.x Let y= (ax^(2) + bx + c)/( px^(2) + qx+ f)

sin^(n)(ax^(2)+bx+c)