If int x^5 e^{-4 x^3} ,d x=frac{1}{48} e^{-4 | vedclass

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(B) Let $I = \int x^5 e^{-4 x^3} \,d x$. Substitute $x^3 = t$,then $3x^2 \,d x = dt$,which implies $x^2 \,d x = \frac{1}{3} dt$. Since $x^5 = x^3 \cdo

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$-4 x^3-1$

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(B) Let $I = \int x^5 e^{-4 x^3} \,d x$. Substitute $x^3 = t$,then $3x^2 \,d x = dt$,which implies $x^2 \,d x = \frac{1}{3} dt$. Since $x^5 = x^3 \cdot x^2$,the integral becomes $I = \frac{1}{3} \int t e^{-4 t} dt$. Using integration by parts,$\int u v \,dt = u \int v \,dt - \int (u' \int v \,dt) dt$,where $u = t$ and $v = e^{-4 t}$. $I = \frac{1}{3} \left( t \cdot \frac{e^{-4 t}}{-4} - \int 1 \cdot \frac{e^{-4 t}}{-4} dt \right)$. $I = \frac{1}{3} \left( -\frac{t e^{-4 t}}{4} + \frac{1}{4} \int e^{-4 t} dt \right)$. $I = \frac{1}{3} \left( -\frac{t e^{-4 t}}{4} - \frac{e^{-4 t}}{16} \right) + c$. $I = -\frac{t e^{-4 t}}{12} - \frac{e^{-4 t}}{48} + c$. $I = \frac{e^{-4 t}}{48} (-4 t - 1) + c$. Substituting $t = x^3$ back,we get $I = \frac{1}{48} e^{-4 x^3} (-4 x^3 - 1) + c$. Comparing this with the given form,$f(x) = -4 x^3 - 1$.

If $\int x^5 e^{-4 x^3} \,d x=\frac{1}{48} e^{-4 x^3} f(x)+c$,where $c$ is a constant of integration,then $f(x)$ is given by

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