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


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Let ![]()
We can write ![]()
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Hence, we can write I2 as

Recall ![]()

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Substituting I1 and I2 in I, we get
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Thus, 