# Find: <span

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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