Q. 16

# 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 – Rate this question :

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