To evaluate int x^3 e^{3x^2 + 5} dx,the simpl | vedclass

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(B) The simplest way is to substitute $3x^2 + 5 = t$. Let $t = 3x^2 + 5$. Then $dt = 6x dx$,which implies $x dx = \frac{dt}{6}$. Also,$x^2 = \frac{t -

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Substitute $3x^2 + 5 = t$

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(B) The simplest way is to substitute $3x^2 + 5 = t$. Let $t = 3x^2 + 5$. Then $dt = 6x dx$,which implies $x dx = \frac{dt}{6}$. Also,$x^2 = \frac{t - 5}{3}$. Substituting these into the integral: $\int x^3 e^{3x^2 + 5} dx = \int x^2 \cdot e^{3x^2 + 5} \cdot x dx = \int \left( \frac{t - 5}{3} \right) e^t \cdot \frac{dt}{6} = \frac{1}{18} \int (t - 5) e^t dt$. $= \frac{1}{18} \left[ \int t e^t dt - 5 \int e^t dt \right] = \frac{1}{18} [ (t e^t - e^t) - 5 e^t ] + c = \frac{1}{18} (t e^t - 6 e^t) + c$. Substituting back $t = 3x^2 + 5$: $= \frac{1}{18} (3x^2 + 5) e^{3x^2 + 5} - \frac{6}{18} e^{3x^2 + 5} + c = \frac{1}{18} (3x^2 + 5) e^{3x^2 + 5} - \frac{1}{3} e^{3x^2 + 5} + c$.

To evaluate $\int x^3 e^{3x^2 + 5} dx$,the simplest way is to

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