Evaluate the following integrals –
Let
Let us assume
We know and derivative of a constant is 0.
⇒ 2x – 5 = λ(2x2-1 – 4 + 0) + μ
⇒ 2x – 5 = λ(2x – 4) + μ
⇒ 2x – 5 = 2λx + μ – 4λ
Comparing the coefficient of x on both sides, we get
2λ = 2 ⇒ λ = 1
Comparing the constant on both sides, we get
μ – 4λ = –5
⇒ μ – 4(1) = –5
⇒ μ – 4 = –5
∴ μ = –1
Hence, we have
Substituting this value in I, we can write the integral as
Let
Now, put x2 – 4x + 3 = t
⇒ (2x – 4)dx = dt (Differentiating both sides)
Substituting this value in I1, we can write
Recall
Let
We can write x2 – 4x + 3 = x2 – 2(x)(2) + 22 – 22 + 3
⇒ x2 – 4x + 3 = (x – 2)2 – 4 + 3
⇒ x2 – 4x + 3 = (x – 2)2 – 1
⇒ x2 – 4x + 3 = (x – 2)2 – 12
Hence, we can write I2 as
Recall
Substituting I1 and I2 in I, we get
Thus,