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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_{}^{} {\sqrt x {e^{\sqrt x }}\;dx = } $
  • A
    $2\sqrt x - {e^{\sqrt x }} - 4\sqrt x \;{e^{\sqrt x }} + c$
  • ✓
    $(2x - 4\sqrt x + 4){e^{\sqrt x }} + c$
  • C
    $(2x + 4\sqrt x + 4){e^{\sqrt x }} + c$
  • D
    $(1 - 4\sqrt x ){e^{\sqrt x }} + c$
✓

Answer

Correct option: B.
$(2x - 4\sqrt x + 4){e^{\sqrt x }} + c$
b
(b) $I = \int_{}^{} {\sqrt x .{e^{\sqrt x }}} dx$. Let $x = {t^2} \Rightarrow dx = 2t\,dt$
$\therefore I = 2\int_{}^{} {{t^2}} \,.\,{e^t}dt$ ==> $I = 2({t^2}.{e^t} - 2t{e^t} + 2{e^t}] + c$
==> $I = \frac{1}{{\sqrt 2 }}\log \left| {\tan \left( {\frac{x}{2} + \frac{{3\pi }}{8}} \right)\,} \right| + c$
i.e., $I = {e^{\sqrt x }}[2x - 4\sqrt x + 4] + c$.

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