Q. 185.0( 2 Votes )

# Find: <span

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 = Rate this question :

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