# Evaluate the integral:

Let, I =

I =

If we assume x2 to be an another variable, we can simplify the integral as derivative of x2 i.e. x is present in numerator.

Let, x2 = u

2x dx = du

x dx = 1/2 du

I =

As,

Let, u – 3 = A(2u + 2) + B

u – 3 = 2Au + 2A + B

On comparing both sides –

We have,

2A = 1 A = 1/2

2A + B = –3 B = –3–2A = –4

Hence,

I =

I =

Let, I1 = and I2 =

Now, I = I1 – 4I2 ….eqn 1

We will solve I1 and I2 individually.

As, I1 =

Let v = u2 + 2u – 4 dv = (2u + 2)du

I1 reduces to

Hence,

I1 = { }

On substituting value of u, we have:

I1 = ….eqn 2

As, I2 = and we don’t have any derivative of function present in denominator.

we will use some special integrals to solve the problem.

As denominator doesn’t have any square root term. So one of the following two integrals will solve the problem.

Now we have to reduce I2 such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of x is seen in denominator.

I2 =

I2 =

Using: a2 + 2ab + b2 = (a + b)2

We have:

I2 =

I2 matches with

I2 = …eqn 3

From eqn 1:

I = I1 – 4I2

Using eqn 2 and eqn 3:

I =

I =

Putting value of u in I:

I =

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