Using integral calculus integrate the function dy/y^2+8y+20
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We have to integrate [1/ (y^2 + 8y + 20) dy]
Int [1/( y^2 + 8y + 20) dy]
=> Int [ 1 / (y^2 + 8y + 16 + 4) dy]
=> Int [ 1/((y + 4)^2 + 2^2) dy]
if u = y + 4 , dy = du
=> Int [ 1/ ( u^2 + 2^2) du]
=> (1/2)*arc tan (u /2) + C
substitute u = y + 4
=> (1/2)*arc tan[(y + 4)/2] + C
The required integral is (1/2)*arc tan[(y + 4)/2] + C
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In other words, we'll have to calculate the indefinite integral of the given function:
Int dy/(y^2+8y+20)
We can re-write the denominator by completing the square:
(y^2+8y+16) - 16 + 20 = (y+4)^2 + 4 = (y+4)^2 + 2^2
Int dy/(y^2+8y+20) = Int dy/[(y+4)^2 + 2^2]
We'll note y + 4 = t => dy = dt
Int dy/[(y+4)^2 + 2^2] = Int dt/(t^2 + 2^2)
Int dt/(t^2 + 2^2) = (1/2)*arctan (t/2) + c
The final result is: Int dy/(y^2+8y+20) = (1/2)*arctan [(y+4)/2]+ c
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