# Evaluate the following integrals – Let Let us assume   We know and derivative of a constant is 0.

x – 3 = λ(2x2-1 + 3 + 0) + μ

x – 3 = λ(2x + 3) + μ

x – 3 = 2λx + 3λ + μ

Comparing the coefficient of x on both sides, we get

2λ = 1 Comparing the constant on both sides, we get

3λ + μ = –3   Hence, we have Substituting this value in I, we can write the integral as    Let Now, put x2 + 3x – 18 = t

(2x + 3)dx = dt (Differentiating both sides)

Substituting this value in I1, we can write  Recall      Let We can write    Hence, we can write I2 as Recall    Substituting I1 and I2 in I, we get Thus, Rate this question :

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