Question 380 of 416MediumMCQ
If \[\int \, \]x5.e-4x3 dx = \[{1 \over {48}}\]e-4x3 f(x) + C, where C is a constant of inegration, then f(x) is equal to -
Solutions · 1
Official solution
\[\int {{x^5}} .{e^{ - 4{x^3}}}\,dx = {1 \over {48}}{e^{ - 4{x^3}}}f\left( x \right) + c\]
Put x3=t
\[3{x^2}\,dx = dt\]
\[\int {{x^3}.{e^{ - 4{x^3}}}.\,{x^2}} dx\]
\[{1 \over 3}\int {t.{e^{ - 4t}}dt} \]
\[{1 \over 3}\left[ {t.{{{e^{ - 4t}}} \over { - 4}} - \int {{{{e^{ - 4t}}} \over { - 4}}dt} } \right]\]
\[ - {{{e^{ - 4t}}} \over {48}}\left[ {4t + 1} \right] + c\]
\[{{ - {e^{ - 4{x^3}}}} \over {48}}\left[ {4{x^3} + 1} \right] + c\]
\[ \therefore \] f(x)=−1−4x3 Answer the question to unlock