Evaluate

Let I =

Dividing 5x^{2} by x^{2} + 4x + 3 we get 5 as quotient and –(20x + 15) as remainder

So, I =

⇒ I = =

⇒ I = 5 (2 – 1) -

⇒ I = 5 – I_{1}

I_{1} =

Adding and subtracting 25 in the numerator

I_{1} =

I_{1} =

Let x^{2} + 4x + 3 = t

(2x + 4)dx = dt

∴ I_{1} =

I_{1} = 10 log t - *[**]*

I_{1} = 10 log t - *[**]*

I_{1} =

I_{1} = 10

I_{1} =

I_{1} =

I_{1} =

I_{1} =

I_{1} =

∵ I = 5 – I_{1}

Substituting I_{1} in I we get

I = 5 –

∴ = 5 –

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