Answer :

Let I =


I =


Let I1 = …(1)


and I2 = …(2)


Thus I = I1 + 3I2 …(3)


I1 can be solved using the substitution method whereas for I2 we will be trying to use the standard formula



Let’s proceed with I2 first –


As I2 =


To get the form present in the formula we need to complete the square. For this, we will be adding 4 and subtracting the same inside the square root.


I2 =


I2 =


Using the formula:


I2 = …(4)


As I1 =


To apply substitution method we need -2x-4 outside square root.


So we proceed as follows:


I1 =


I1 =


Let, 3 – 4x – x2 = u


du = (-2x – 4) dx


I1 = {using equation 2}


I1 =


I1 = …(5)


From equation 3,4 and 5:


I =


I =


Hence,


I =


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