Medium MCQ +4 / -1 PYQ · JEE Mains 2023

The integral $16\int\limits_1^2 {{{dx} \over {{x^3}{{\left( {{x^2} + 2} \right)}^2}}}}$ is equal to

  1. A ${{11} \over {12}} + {\log _e}4$
  2. B ${{11} \over 6} + {\log _e}4$
  3. C ${{11} \over {12}} - {\log _e}4$
  4. D ${{11} \over 6} - {\log _e}4$ Correct answer

Solution

$I=\int \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}$ <br/><br/>$=\frac{1}{4} \int \frac{x}{x^{2}+2} d x+\frac{1}{4} \int \frac{x}{\left(x^{2}+2\right)^{2}}-\frac{1}{4} \int \frac{d x}{x}+\frac{1}{4} \int \frac{d x}{x^{3}}$<br/><br/> $=\frac{1}{8} \ln \left(x^{2}+2\right)-\frac{\ln x}{4}-\frac{1}{8\left(x^{2}+2\right)}-\frac{1}{8 x^{3}}$ <br/><br/> Now, $16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}=2 \ln 6-2 \ln 3-4 \ln 2+\frac{11}{6}$ <br/><br/>$=\frac{11}{6}-\ln 4$

About this question

Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals

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