# Using integral calculus integrate the function dy/y^2+8y+20

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### 2 Answers

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**

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**