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,