Question

Find : `int(x+3)sqrt(3-4x-x^2dx)`

Answer 1

`I=int(x+3)sqrt(3-4x-x^2)dx`

Let (x+3)= `Ad/dx(3-4x-x^2)+B `

⇒ x+3=A(-2x-4)+B

⇒ x+3 = -2Ax-4A +B

∴ -2A=1

⇒A =`1/2`

-4A+B=3

⇒ `-4(-1/2)+B=3`

⇒ B =1

`:.I=int[-1/2d/dx(3-4x-x^2)+1]sqrt(3-4x-x^2dx)`

`=1/2intd/dx(3-4x-x^2)sqrt(3-4x-x^2)dx+intsqrt(3-4x-x^2-4+4)dx`

`=1/2((3-4x-x^2)^(3/2)/(3/2))+intsqrt(7-(x+2)^2)dx`

 `=(3-4x-x^2)^(3/2)/3+(x+2)/2sqrt(7-(x+2)^2)+7/2sin^(-1)((x+2)/sqrt7)+C`