Question
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Answer :
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### Problem i) Integrate \(\frac{1}{x^2 - 6x + 11}\) dx #### Step 1: Complete the Square First, let's complete the square for the denominator \(x^2 - 6x + 11\): \[ x^2 - 6x + 11 = (x^2 - 6x + 9) + 2 = (x - 3)^2 + 2 \] So, the integral becomes: \[ \int \frac{1}{(x - 3)^2 + 2} \, dx \] #### Step 2: Substitute Let \(u = x - 3\), then \(du = dx\). The integral becomes: \[ \int \frac{1}{u^2 + 2} \, du \] #### Step 3: Recognize the Standard Integral This is a standard integral of the form: \[ \int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C \] Here, \(a^2 = 2\), so \(a = \sqrt{2}\). Thus: \[ \int \frac{1}{u^2 + 2} \, du = \frac{1}{\sqrt{2}} \arctan\left(\frac{u}{\sqrt{2}}\right) + C \] #### Step 4: Substitute Back Replace \(u\) with \(x - 3\): \[ \frac{1}{\sqrt{2}} \arctan\left(\frac{x - 3}{\sqrt{2}}\right) + C \] #### Final Answer: \[ \boxed{\frac{1}{\sqrt{2}} \arctan\left(\frac{x - 3}{\sqrt{2}}\right) + C} \] --- ### Problem ii) Integrate \(\frac{x^2 + 1}{(2x + 1)(x - 1)(x + 1)} \, dx\) #### Step 1: Partial Fraction Decomposition We want to decompose: \[ \frac{x^2 + 1}{(2x + 1)(x - 1)(x + 1)} = \frac{A}{2x + 1} + \frac{B}{x - 1} + \frac{C}{x + 1} \] Multiply both sides by the denominator \((2x + 1)(x - 1)(x + 1)\): \[ x^2 + 1 = A(x - 1)(x
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