Evaluate

Let I =

Dividing 5x² by x² + 4x + 3 we get 5 as quotient and –(20x + 15) as remainder

So, I =

⇒ I = =

⇒ I = 5 (2 – 1) -

⇒ I = 5 – I₁

I₁ =

Adding and subtracting 25 in the numerator

I₁ =

I₁ =

Let x² + 4x + 3 = t

(2x + 4)dx = dt

∴ I₁ =

I₁ = 10 log t - []

I₁ = 10 log t - []

I₁ =

I₁ = 10

I₁ =

I₁ =

I₁ =

I₁ =

I₁ =

∵ I = 5 – I₁

Substituting I₁ in I we get

I = 5 –

∴ = 5 –