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STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.1 inderfinite integral4 Marks
MCQ
A primitive of $f(x) = \frac{x}{{1 + {x^2}}}$  $dx$=
  • A
    ${\log _e}({x^2} + 1)$
  • B
    $x{\tan ^{ - 1}}x$
  • ✓
    $\frac{{{{\log }_e}({x^2} + 1)}}{2}$
  • D
    $\frac{1}{2}x{\tan ^{ - 1}}x$
✓

Answer

Correct option: C.
$\frac{{{{\log }_e}({x^2} + 1)}}{2}$
c
(c)$f(x) = \frac{x}{{1 + {x^2}}}$, $\therefore \,\,\,I = \int_{}^{} {f(x)} = \int_{}^{} {\frac{x}{{1 + {x^2}}}\,dx} $
Put $1 + {x^2} = t$

$\Rightarrow 2x\,dx = dt \Rightarrow x\,dx = dt2$
$\therefore \,\,\,I = \frac{1}{2}\int_{}^{} {\frac{{dt}}{t} = \frac{1}{2}\log t + c} $; $I = \frac{1}{2}\log (1 + {x^2}) + c$.

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