Answer:
[tex]\begin{gathered} \large\qquad\qquad\boxed{ \sf{ \: \bold{ \: \: (1) \: \: - {4x}^{3} - 1 \: \: }}} \\ \end{gathered} [/tex]
Step-by-step explanation:
[tex] \underline{ \sf{ \pmb{Given \: integral }\: is : }}[/tex]
[tex]\begin{gathered} \quad \hookrightarrow \: \boxed{\bold{\displaystyle{ \bold{\int\bold{ {x}^{5} {e}^{ - {4x}^{3} } \: dx \: = \: \dfrac{1}{48}{e}^{ - {4x}^{3} } + c }}}}}\\ \end{gathered} [/tex]
[tex] \sf{Now, \: Consider} : [/tex]
[tex]\begin{gathered} \qquad \qquad \hookrightarrow\sf \:\displaystyle\int\sf {x}^{5} {e}^{ - {4x}^{3} } \: dx \: \\ \end{gathered} [/tex]
[tex] \sf{can \: be \: rewritten \: as} : [/tex]
[tex]\begin{gathered}\sf \qquad \qquad \hookrightarrow \: \displaystyle\int\sf {x}^{3} {e}^{ - {4x}^{3} } \: {x}^{2} \: dx \: \\ \end{gathered} [/tex]
[tex] \sf{can \: be \: rewritten \: as} : [/tex]
[tex]\begin{gathered} \qquad\sf \implies \: \dfrac{1}{3} \displaystyle\int\sf {x}^{3} {e}^{ - {4x}^{3} } \: 3{x}^{2} \: dx \: \\ \end{gathered} [/tex]
[tex] \sf{Now, \: Substitute} : [/tex]
[tex]\begin{gathered}\sf \qquad \implies \: {x}^{3} = y \\ \end{gathered}[/tex]
[tex]\begin{gathered}\sf \qquad\implies \sf \: 3{x}^{2} \: dx = dy \\ \end{gathered} [/tex]
So, on substituting these values in above integral, we get:
[tex]\begin{gathered}\sf \qquad \implies \: \dfrac{1}{3} \displaystyle\int\sf y {e}^{ - 4y } \: dy \: \\ \end{gathered} [/tex]
[tex] \sf{On \: \pmb{ integrating \: by \: parts}, \: we \: \pmb{get}} : [/tex]
[tex]\begin{gathered}\sf \implies \: \dfrac{1}{3}\left[ y\displaystyle\int\sf {e}^{ - 4y } \: dy - \displaystyle\int\sf \bigg(\dfrac{d}{dy}y\displaystyle\int\sf {e}^{ - 4y} \: dy \bigg)dy \:\right] \\ \end{gathered}[/tex]
[tex]\begin{gathered}\sf \implies \: \dfrac{1}{3}\left[y\dfrac{ {e}^{ - 4y} }{ - 4} - \displaystyle\int\sf \bigg(\dfrac{ {e}^{ - 4y} }{ - 4}\bigg)dy \:\right] \\ \end{gathered} [/tex]
[tex]\begin{gathered}\sf \implies \: \dfrac{1}{3}\left[y\dfrac{ {e}^{ - 4y} }{ - 4} -\dfrac{ {e}^{ - 4y} }{16} \:\right] + c \\ \end{gathered} [/tex]
[tex]\begin{gathered}\sf \implies \: - \dfrac{ {e}^{ - 4y} }{48}(4y + 1) + c\\ \end{gathered}[/tex]
[tex]\begin{gathered}\sf \implies \: - \dfrac{{e}^{ - {4x}^{3} } }{48}(4 {x}^{3} + 1) + c\\ \end{gathered} [/tex]
[tex]\begin{gathered}\sf \implies \: \dfrac{{e}^{ - {4x}^{3} } }{48}( - 4 {x}^{3} - 1) + c\\ \end{gathered} [/tex]
[tex] \large \sf{Thus,}[/tex]
[tex]\begin{gathered}\sf\implies \sf \:\displaystyle\int\sf {x}^{5} {e}^{ - {4x}^{3} } \: dx \: = \: \dfrac{{e}^{ - {4x}^{3} } }{48}( - 4 {x}^{3} - 1) + c \\ \end{gathered} [/tex]
[tex] \sf{ \pmb{But} \: is \: \pmb{ given} \: that,}[/tex]
[tex]\begin{gathered}\sf \implies \:\displaystyle\int\sf {x}^{5} {e}^{ - {4x}^{3} } \: dx \: = \: \dfrac{1}{48}{e}^{ - {4x}^{3} } + c \\ \end{gathered} [/tex]
[tex] \underline{ \sf{ \pmb{So}, \: on \: \pmb{comparing} \: we \: \pmb{get} : }}[/tex]
[tex]\begin{gathered} \\ \qquad\color{yellow}{\bigstar}\\ \sf \qquad \color{yellow}{ \bigstar} - \: \boxed{\bold{ \green{ \pmb{ \: \: \: f(x) = - {4x}^{3} - 1 \: \: \: }}}} \: - \bigstar\\ \qquad\color{yellow}{\bigstar} \end{gathered} [/tex]
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[tex] \large\underline{ \sf{ \pmb{Basic \: Concept \: Used : }}} \\ [/tex]
[tex] \underline{ \sf{Integration \: by \: parts : }}[/tex]
[tex]\begin{gathered}\boxed{\sf\int u \cdot v\, dx = u\int v \, dx- \int \left[\dfrac{du}{dx}\int v\, dx\right] dx} \\ \end{gathered}[/tex]
Here, u and v are generally chosen according to the word ILATE.
[tex] \sf{In \: the \: word \: ILATE,}[/tex]
[tex] \qquad\sf{I - \pmb{ stands \: for \: inverse \: functions.}} \\ [/tex]
[tex]\qquad \sf{L - \pmb{ stands \: for \: logarithmic \: functions.}} \\ [/tex]
[tex] \qquad\sf{A - \pmb{stands \: for \: algebraic \: functions.}} \\ [/tex]
[tex]\qquad \sf{T - \pmb{stands \: for \: trigonometric \: functions.}} \\ [/tex]
[tex] \qquad\sf{E - \pmb{stands \: for \: exponential \: functions.}}[/tex]
