Q. 6

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

x – 2 = λ(2 × 2x2-1 – 6 – 0) + μ

x – 2 = λ(4x – 6) + μ

x – 2 = 4λx + μ – 6λ

Comparing the coefficient of x on both sides, we get

4λ = 1 Comparing the constant on both sides, we get

μ – 6λ = –2   Hence, we have Substituting this value in I, we can write the integral as    Let Now, put 2x2 – 6x + 5 = t

(4x – 6)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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