A) \[\frac{1}{6}{{e}^{3{{t}^{2}}}}+c\]
B) \[-\frac{1}{5}{{e}^{3{{t}^{2}}}}+c\]
C) \[\frac{1}{6}{{e}^{-3{{t}^{2}}}}+c\]
D) \[-\frac{1}{6}{{e}^{-3{{t}^{2}}}}+c\]
Correct Answer: D
Solution :
\[\int{\frac{t}{{{e}^{3{{t}^{2}}}}}}dt\] let \[3{{t}^{2}}=x,6t\,dt=dx\] \[\Rightarrow \] \[t\,dt=\frac{1}{6}dx\] \[\therefore \] \[\int{\frac{t}{{{e}^{3{{t}^{2}}}}}}dt=\int{\frac{1}{{{e}^{x}}}}\frac{1}{6}dx=\frac{1}{6}\int{{{e}^{-x}}}dx\] \[=-\frac{{{e}^{-x}}}{6}+c\] \[=-\frac{1}{6}{{e}^{-3{{t}^{2}}}}+c\]
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