$$ J = \int \frac { 5 x } { e ^ { 3 - 2 x ^ { 2 } } } d x $$

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Calculate the integral $$ \int \frac{5x}{e^{3-2x^{2}}}dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int \frac{5x}{e^{3-2x^{2}}} dx$$
- step1: Simplify the expression:
  $$\int 5xe^{-3+2x^{2}} dx$$
- step2: Use properties of integrals:
  $$5	imes \int xe^{-3+2x^{2}} dx$$
- step3: Use the substitution $$dx=\frac{1}{4x} dt$$ to transform the integral$$:$$
  $$5	imes \int xe^{-3+2x^{2}}	imes \frac{1}{4x} dt$$
- step4: Simplify:
  $$5	imes \int \frac{e^{-3+2x^{2}}}{4} dt$$
- step5: Use the substitution $$t=2x^{2}$$ to transform the integral$$:$$
  $$5	imes \int \frac{e^{-3+t}}{4} dt$$
- step6: Rewrite the expression:
  $$5	imes \int \frac{1}{4}e^{-3+t} dt$$
- step7: Use properties of integrals:
  $$5	imes \frac{1}{4}	imes \int e^{-3+t} dt$$
- step8: Multiply:
  $$\frac{5}{4}	imes \int e^{-3+t} dt$$
- step9: Use the substitution $$dt=1 dv$$ to transform the integral$$:$$
  $$\frac{5}{4}	imes \int e^{-3+t}	imes 1 dv$$
- step10: Simplify:
  $$\frac{5}{4}	imes \int e^{-3+t} dv$$
- step11: Use the substitution $$v=-3+t$$ to transform the integral$$:$$
  $$\frac{5}{4}	imes \int e^{v} dv$$
- step12: Evaluate the integral:
  $$\frac{5}{4}e^{v}$$
- step13: Multiply the terms:
  $$\frac{5e^{v}}{4}$$
- step14: Substitute back:
  $$\frac{5e^{-3+t}}{4}$$
- step15: Substitute back:
  $$\frac{5e^{-3+2x^{2}}}{4}$$
- step16: Add the constant of integral C:
  $$\frac{5e^{-3+2x^{2}}}{4} + C, C \in \mathbb{R}$$
The integral of $$ \frac{5x}{e^{3-2x^{2}}} $$ with respect to $$ x $$ is $$ \frac{5e^{-3+2x^{2}}}{4} + C $$, where $$ C $$ is an arbitrary constant.
The integral of $$ \frac{5x}{e^{3-2x^{2}}} $$ with respect to $$ x $$ is $$ \frac{5e^{-3+2x^{2}}}{4} + C $$.

\( J = \int \frac { 5 x } { e ^ { 3 - 2 x ^ { 2 } } } d x \)

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