Solution 2.3:1a

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The formula for integration by parts reads
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{{Displayed math||<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}
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<center> [[Bild:2_3_1a-2(2).gif]] </center>
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where <math>F(x)</math> is a primitive function of <math>f(x)</math> and <math>g'(x)</math> is a derivative of <math>g(x)</math>.
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If we are to use integration by parts, the integrand has to be divided up into two factors, a factor <math>f(x)</math> which we integrate and a factor <math>g(x)</math> which we differentiate. It is only when the product <math>F(x)g'(x)</math>
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becomes simpler than <math>f(x)g(x)</math> that there is any point in integrating by parts.
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In the integral
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{{Displayed math||<math>\int 2xe^{-x}\,dx\,,</math>}}
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it can seem appropriate to choose <math>f(x)=e^{-x}</math> and <math>g(x) = 2x</math>, because then <math>g'(x) = 2</math> and we have only <math>F(x) = -e^{-x}</math> left,
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{{Displayed math||<math>\begin{align}
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\int 2x\cdot e^{-x}\,dx
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&= 2x\cdot \bigl(-e^{-x}\bigr) - \int 2\cdot \bigl(-e^{-x}\bigr)\,dx\\[5pt]
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&= -2xe^{-x} + 2\int e^{-x}\,dx\,\textrm{.}
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\end{align}</math>}}
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It remains only to integrate <math>e^{-x}</math> and we are finished,
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{{Displayed math||<math>\begin{align}
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\phantom{\int 2x\cdot e^{-x}\,dx}{}
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&= \rlap{-2xe^{-x} + 2\bigl(-e^{-x}\bigr) + C}\phantom{2x\cdot \bigl(-e^{-x}\bigr) - \int 2\cdot \bigl(-e^{-x}\bigr)\,dx}\\[5pt]
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&= -2xe^{-x} - 2e^{-x} + C\\[5pt]
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&= -2(x+1)e^{-x} + C\,\textrm{.}
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\end{align}</math>}}

Current revision

The formula for integration by parts reads

\displaystyle \int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,

where \displaystyle F(x) is a primitive function of \displaystyle f(x) and \displaystyle g'(x) is a derivative of \displaystyle g(x).

If we are to use integration by parts, the integrand has to be divided up into two factors, a factor \displaystyle f(x) which we integrate and a factor \displaystyle g(x) which we differentiate. It is only when the product \displaystyle F(x)g'(x) becomes simpler than \displaystyle f(x)g(x) that there is any point in integrating by parts.

In the integral

\displaystyle \int 2xe^{-x}\,dx\,,

it can seem appropriate to choose \displaystyle f(x)=e^{-x} and \displaystyle g(x) = 2x, because then \displaystyle g'(x) = 2 and we have only \displaystyle F(x) = -e^{-x} left,

\displaystyle \begin{align}

\int 2x\cdot e^{-x}\,dx &= 2x\cdot \bigl(-e^{-x}\bigr) - \int 2\cdot \bigl(-e^{-x}\bigr)\,dx\\[5pt] &= -2xe^{-x} + 2\int e^{-x}\,dx\,\textrm{.} \end{align}

It remains only to integrate \displaystyle e^{-x} and we are finished,

\displaystyle \begin{align}

\phantom{\int 2x\cdot e^{-x}\,dx}{} &= \rlap{-2xe^{-x} + 2\bigl(-e^{-x}\bigr) + C}\phantom{2x\cdot \bigl(-e^{-x}\bigr) - \int 2\cdot \bigl(-e^{-x}\bigr)\,dx}\\[5pt] &= -2xe^{-x} - 2e^{-x} + C\\[5pt] &= -2(x+1)e^{-x} + C\,\textrm{.} \end{align}

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