MCQ
$\int\left(x+\frac{1}{x}\right)^3 d x=$
  • A
    $\frac{1}{4}\left(x+\frac{1}{x}\right)^4+c$
  • $\frac{x^4}{4}+\frac{3 x^2}{2}+3 \log x-\frac{1}{2 x^2}+c$
  • C
    $\frac{x^4}{4}+\frac{3 x^2}{2}+3 \log x+\frac{1}{x^2}+c$
  • D
    $\left(x-x^{-1}\right)^3+c$

Answer

Correct option: B.
$\frac{x^4}{4}+\frac{3 x^2}{2}+3 \log x-\frac{1}{2 x^2}+c$
$\int\left(x+\frac{1}{x}\right)^3 d x=\frac{ x ^4}{4}+\frac{3 x ^2}{2}+3 \log x -\frac{1}{2 x ^2}+ c .$
Explanation:
$\left(x+\frac{1}{x}\right)^3=x^3+3 x+\frac{3}{x}+\frac{1}{x^3}$
$\therefore \int\left(x+\frac{1}{x}\right)^3 d x=\int\left(x^3+3 x+\frac{3}{x}+\frac{1}{x^3}\right) d x$
$=\frac{x^4}{4}+\frac{3 x^2}{2}+3 \log x-\frac{1}{2 x^2}+c$

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