int {{x^5}{e^{{x^2}}}} dx = | vedclass

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Question

(A) Let $I = \int {{x^5}{e^{{x^2}}}} dx$. Substitute ${x^2} = t$,then $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$. Also,${x^4} = {t^2}$. Substi

Vedclass · Accepted Answer

$\frac{1}{2}{x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c$

Vedclass · Answer

(A) Let $I = \int {{x^5}{e^{{x^2}}}} dx$. Substitute ${x^2} = t$,then $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$. Also,${x^4} = {t^2}$. Substituting these into the integral: $I = \int {{x^4} \cdot {e^{{x^2}}} \cdot x dx} = \int {{t^2}{e^t} \cdot \frac{1}{2} dt} = \frac{1}{2} \int {{t^2}{e^t} dt}$. Using integration by parts,$\int {u v dt} = u \int v dt - \int (u' \int v dt) dt$: $I = \frac{1}{2} [ {t^2}{e^t} - \int {2t{e^t} dt} ] = \frac{1}{2} {t^2}{e^t} - \int {t{e^t} dt}$. Applying integration by parts again to $\int {t{e^t} dt}$: $I = \frac{1}{2} {t^2}{e^t} - [ t{e^t} - \int {e^t dt} ] = \frac{1}{2} {t^2}{e^t} - t{e^t} + {e^t} + c$. Substituting $t = {x^2}$ back: $I = \frac{1}{2} {x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c$.

$\int {{x^5}{e^{{x^2}}}} dx = $

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