NCERT Class 12 Maths · Chapter 7 Integrals · Exercise 7.9 Question 6

✅ Step-by-Step Solution

1. Work on each open interval where the original integrand is real and defined.
2. Complete the square: 17/4−(x−1/2)²; set u=(2x−1)/√17.
3. An antiderivative is F(x)=\(\frac{\sqrt{17} \log{\left(\left|{\frac{2 x - 1 + \sqrt{17}}{- 2 x + 1 + \sqrt{17}}}\right| \right)}}{17}\).
4. At the upper and lower limits, F(b)=\(\frac{\sqrt{17} \log{\left(\frac{3 + \sqrt{17}}{-3 + \sqrt{17}} \right)}}{17}\) and F(a)=\(\frac{\sqrt{17} \log{\left(\frac{-1 + \sqrt{17}}{1 + \sqrt{17}} \right)}}{17}\).
5. By the fundamental theorem (using one-sided limits for any integrable endpoint), I=F(b)−F(a)=\(\log{\left(\left(\frac{6 \sqrt{17} + 26}{9 - \sqrt{17}}\right)^{\frac{\sqrt{17}}{17}} \right)}\).