MCQ
$\int_{}^{} {\frac{{5({x^6} + 1)}}{{{x^2} + 1}}dx = } $
  • A
    $5({x^7} + x){\tan ^{ - 1}}x + c$
  • ${x^5} - \frac{5}{3}{x^3} + 5x + c$
  • C
    $3{x^4} - 5{x^2} + 15x + c$
  • D
    $5{\tan ^{ - 1}}({x^2} + 1) + \log ({x^2} + 1) + c$

Answer

Correct option: B.
${x^5} - \frac{5}{3}{x^3} + 5x + c$
b
(b)$\int_{}^{} {\frac{{5({x^6} + 1)}}{{{x^2} + 1}}\,dx = \int_{}^{} {\frac{{5({x^2} + 1)({x^4} - {x^2} + 1)}}{{({x^2} + 1)}}\,dx} } $
$ = \int_{}^{} {5({x^4} - {x^2} + 1)\,dx = {x^5} - \frac{5}{3}{x^3} + 5x + c.} $

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