MCQ
જો $A = \int\limits_0^1 {\frac{{{e^t}}}{{t + 1}}dt,} $તો$\int\limits_0^1 {\frac{{t{e^{{t^2}}}}}{{{t^2} + 1}}dt = .......} $
- A$A$
- B$2A$
- ✓$\frac{1}{2}A$
- D${A^2}$
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${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$
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