# Evaluate  [CBSE 2011]

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 –

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