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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
$\int_{}^{} {x\sqrt {1 + {x^2}} } \;dx = $
  • A
    $\frac{{1 + 2{x^2}}}{{\sqrt {1 + {x^2}} }} + c$
  • B
    $\sqrt {1 + {x^2}} + c$
  • C
    $3{(1 + {x^2})^{3/2}} + c$
  • ✓
    $\frac{1}{3}{(1 + {x^2})^{3/2}} + c$
✓

Answer

Correct option: D.
$\frac{1}{3}{(1 + {x^2})^{3/2}} + c$
d
(d) Put $1 + {x^2} = t \Rightarrow x\,dx = \frac{{dt}}{2}$ It reduces to $\frac{1}{2}\int_{}^{} {{t^{1/2}}dt} = \frac{1}{2} \times \frac{{{t^{3/2}}}}{{3/2}} = \frac{1}{3}{(1 + {x^2})^{3/2}} + c.$

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