text: # Integrate x^2e^(-1x)
  
  ## \int x^2 e^{\left(-1\right)\cdot x}dx
  
  Go!
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  $-x^2e^{-x}-2e^{-x}-2xe^{-x}+C_0$
  
  ## Step by step solution
  
  Problem
  
  $\int x^2 e^{\left(-1\right)\cdot x}dx$
  1
  
  Use the integration by parts theorem to calculate the integral $\int x^2e^{-x}dx$, using the following formula
  
  $\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
  2
  
  First, identify $u$ and calculate $du$
  
  $\begin{matrix}\displaystyle{u=x^2}\\ \displaystyle{du=2xdx}\end{matrix}$
  3
  
  Now, identify $dv$ and calculate $v$
  
  $\begin{matrix}\displaystyle{dv=e^{-x}dx}\\ \displaystyle{\int dv=\int e^{-x}dx}\end{matrix}$
  4
  
  Solve the integral
  
  $v=\int e^{-x}dx$
  5
  
  Solve the integral $\int e^{-x}dx$ applying u-substitution. Let $u$ and $du$ be
  
  $\begin{matrix}u=-x \\ du=-1dx\end{matrix}$
  6
  
  Isolate $dx$ in the previous equation
  
  $\frac{du}{-1}=dx$
  7
  
  Substituting $u$ and $dx$ in the integral
  
  $\int-e^udu$
  8
  
  Taking the constant out of the integral
  
  $-\int e^udu$
  9
  
  The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$
  
  $-e^u$
  10
  
  Substitute $u$ back for it's value, $-x$
  
  $-e^{-x}$
  11
  
  Now replace the values of $u$, $du$ and $v$ in the last formula
  
  $2\int xe^{-x}dx-x^2e^{-x}$
  12
  
  Use the integration by parts theorem to calculate the integral $\int xe^{-x}dx$, using the following formula
  
  $\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
  13
  
  First, identify $u$ and calculate $du$
  
  $\begin{matrix}\displaystyle{u=x}\\ \displaystyle{du=dx}\end{matrix}$
  14
  
  Now, identify $dv$ and calculate $v$
  
  $\begin{matrix}\displaystyle{dv=e^{-x}dx}\\ \displaystyle{\int dv=\int e^{-x}dx}\end{matrix}$
  15
  
  Solve the integral
  
  $v=\int e^{-x}dx$
  16
  
  Solve the integral $\int e^{-x}dx$ applying u-substitution. Let $u$ and $du$ be
  
  $\begin{matrix}u=-x \\ du=-1dx\end{matrix}$
  17
  
  Isolate $dx$ in the previous equation
  
  $\frac{du}{-1}=dx$
  18
  
  Substituting $u$ and $dx$ in the integral
  
  $2\int x-e^udx-x^2e^{-x}$
  19
  
  Taking the constant out of the integral
  
  $2\int-xe^udx-x^2e^{-x}$
  20
  
  The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$
  
  $2\int-xe^udx-x^2e^{-x}$
  21
  
  Substitute $u$ back for it's value, $-x$
  
  $2\int-xe^udx-x^2e^{-x}$
  22
  
  Now replace the values of $u$, $du$ and $v$ in the last formula
  
  $2\left(\int e^{-x}dx-xe^{-x}\right)-x^2e^{-x}$
  23
  
  Solve the integral $\int e^{-x}dx$ applying u-substitution. Let $u$ and $du$ be
  
  $\begin{matrix}u=-x \\ du=-1dx\end{matrix}$
  24
  
  Isolate $dx$ in the previous equation
  
  $\frac{du}{-1}=dx$
  25
  
  Substituting $u$ and $dx$ in the integral
  
  $2\left(\int-e^udu-xe^{-x}\right)-x^2e^{-x}$
  26
  
  Taking the constant out of the integral
  
  $2\left(-xe^{-x}-\int e^udu\right)-x^2e^{-x}$
  27
  
  The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$
  
  $2\left(-xe^{-x}-e^u\right)-x^2e^{-x}$
  28
  
  Substitute $u$ back for it's value, $-x$
  
  $2\left(-xe^{-x}-e^{-x}\right)-x^2e^{-x}$
  29
  
  Multiply $\left(-xe^{-x}+-e^{-x}\right)$ by $2$
  
  $-x^2e^{-x}-2e^{-x}-2xe^{-x}$
  30
  
  $-x^2e^{-x}-2e^{-x}-2xe^{-x}+C_0$
  
  $-x^2e^{-x}-2e^{-x}-2xe^{-x}+C_0$
  
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