Evaluate the following integrals –

Let


Let us assume




We know and derivative of a constant is 0.


x + 3 = λ(0 – 4 – 2x2-1) + μ


x + 3 = λ(–4 – 2x) + μ


x + 3 = –2λx + μ – 4λ


Comparing the coefficient of x on both sides, we get


–2λ = 1


Comparing the constant on both sides, we get


μ – 4λ = 3



μ + 2 = 3


μ = 1


Hence, we have


Substituting this value in I, we can write the integral as






Let


Now, put 3 – 4x – x2 = t


(–4 – 2x)dx = dt (Differentiating both sides)


Substituting this value in I1, we can write




Recall







Let


We can write 3 – 4x – x2 = –(x2 + 4x – 3)


3 – 4x – x2 = –[x2 + 2(x)(2) + 22 – 22 – 3]


3 – 4x – x2 = –[(x + 2)2 – 4 – 3]


3 – 4x – x2 = –[(x + 2)2 – 7]


3 – 4x – x2 = 7 – (x + 2)2



Hence, we can write I2 as



Recall




Substituting I1 and I2 in I, we get



Thus,


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