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Class 12 Integrals · Long Answer3 Marksindefinite_integration

Class 12 Maths · Integrals · Long Answer #161

Evaluate \(\int \frac{\left(x^{2} + 1\right) \left(x^{2} + 4\right)}{\left(x^{2} - 5\right) \left(x^{2} + 3\right)}\,dx\). Work on each connected open interval where the real integrand is defined.

✅ Step-by-Step Solution

Final Answer: \(x + \frac{27 \sqrt{5} \log{\left(\left|{\frac{x - \sqrt{5}}{x + \sqrt{5}}}\right| \right)}}{40} + \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3} x}{3} \right)}}{12}\) + C
  1. Polynomial division gives the integrand 1+(7x²+19)/[(x²+3)(x²−5)].
  2. Set the remainder to A/(x²+3)+B/(x²−5). Solve A+B=7,−5A+3B=19 to obtain A=1/4,B=27/4.
  3. Integrate 1, the arctangent term (1/4)/(x²+3), and the logarithmic term (27/4)/(x²−5).
  4. Therefore the required integral is \(x + \frac{27 \sqrt{5} \log{\left(\left|{\frac{x - \sqrt{5}}{x + \sqrt{5}}}\right| \right)}}{40} + \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3} x}{3} \right)}}{12}\) + C.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Polynomial division gives the integrand 1+(7x²+19)/[(x²+3)(x²−5)].
  • [1 mark] Set the remainder to A/(x²+3)+B/(x²−5). Solve A+B=7,−5A+3B=19 to obtain A=1/4,B=27/4.
  • [1 mark] Integrate 1, the arctangent term (1/4)/(x²+3), and the logarithmic term (27/4)/(x²−5).
  • [0.5 marks] Therefore the required integral is \(x + \frac{27 \sqrt{5} \log{\left(\left|{\frac{x - \sqrt{5}}{x + \sqrt{5}}}\right| \right)}}{40} + \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3} x}{3} \right)}}{12}\) + C.

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