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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{{dx}}{{1 + x + {x^2} + {x^3}}} = } $
  • ✓
    $\log \sqrt {1 + x} - \frac{1}{2}\log \sqrt {1 + {x^2}} + \frac{1}{2}{\tan ^{ - 1}}x + c$
  • B
    $\log \sqrt {1 + x} - \log \sqrt {1 + {x^2}} + {\tan ^{ - 1}}x + c$
  • C
    $\log \sqrt {1 + {x^2}} - \log \sqrt {1 + x} + \frac{1}{2}{\tan ^{ - 1}}x + c$
  • D
    $\log \sqrt {1 + x} + {\tan ^{ - 1}}x + \log \sqrt {1 + {x^2}} + c$
✓

Answer

Correct option: A.
$\log \sqrt {1 + x} - \frac{1}{2}\log \sqrt {1 + {x^2}} + \frac{1}{2}{\tan ^{ - 1}}x + c$
a
(a)$\int_{}^{} {\frac{{dx}}{{1 + x + {x^2} + {x^3}}} = \int_{}^{} {\frac{{dx}}{{(1 + x)(1 + {x^2})}}} } $
$ = \frac{1}{2}\int_{}^{} {\frac{1}{{1 + {x^2}}}\,dx} + \frac{1}{2}\int_{}^{} {\frac{1}{{1 + x}}\,dx} - \frac{1}{2}\int_{}^{} {\frac{x}{{1 + {x^2}}}\,dx} $
$ = \frac{1}{2}{\tan ^{ - 1}}x + \log \sqrt {1 + x} - \frac{1}{2}\log \sqrt {1 + {x^2}} + c$.

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