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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx = } $
  • A
    $\log (1 + {x^2}) + c$
  • B
    $\log {e^{{{\tan }^{ - 1}}x}} + c$
  • ✓
    ${e^{{{\tan }^{ - 1}}x}} + c$
  • D
    ${\tan ^{ - 1}}{e^{{{\tan }^{ - 1}}x}} + c$
✓

Answer

Correct option: C.
${e^{{{\tan }^{ - 1}}x}} + c$
c
(c) Putting $t = {\tan ^{ - 1}}x \Rightarrow dt = \frac{1}{{1 + {x^2}}}\,dx,$ we get
$\int_{}^{} {\frac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}\,dx} = \int_{}^{} {{e^t}dt} $$ = {e^t} + c = {e^{{{\tan }^{ - 1}}x}} + c.$

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