Lösung 2.2:3e

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If we differentiate the denominator in the integrand
If we differentiate the denominator in the integrand
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{{Displayed math||<math>(x^2+1)' = 2x</math>}}
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<math>3x=\frac{3}{2}\centerdot 2x=\frac{3}{2}\left( x^{2}+1 \right)^{\prime }</math>
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we obtain almost the same expression as in the numerator; there is a constant 2 which is different. We therefore rewrite the numerator as
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{{Displayed math||<math>3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',</math>}}
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we obtain almost the same expression as in the numerator; there is a constant
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<math>\text{2}</math>
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which is different. We therefore rewrite the numerator as
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<math>3x=\frac{3}{2}\centerdot 2x=\frac{3}{2}\centerdot \left( x^{2}+1 \right)^{\prime }</math>,
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so the integral can be written as
so the integral can be written as
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{{Displayed math||<math>\int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,,</math>}}
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<math>\int{\frac{\frac{3}{2}}{x^{2}+1}}\centerdot \left( x^{2}+1 \right)^{\prime }\,dx</math>,
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and we see that the substitution <math>u=x^2+1</math> can be used to simplify the integral,
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and we see that the substitution
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<math>u=x^{2}+1</math>
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can be used to simplify the integral:
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<math>\begin{align}
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& \int{\frac{3x}{x^{2}+1}}\,dx=\left\{ \begin{matrix}
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u=x^{2}+1 \\
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du=\left( x^{2}+1 \right)^{\prime }\,dx=2x\,dx \\
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\end{matrix} \right\} \\
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& =\frac{3}{2}\int{\frac{\,du}{u}}=\frac{3}{2}\ln \left| u \right|+C \\
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& =\frac{3}{2}\ln \left| x^{2}+1 \right|+C \\
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& =\frac{3}{2}\ln \left( x^{2}+1 \right)+C \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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\int \frac{3x}{x^2+1}\,dx
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&= \left\{ \begin{align}
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u &= x^2+1\\[5pt]
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du &= (x^2+1)'\,dx = 2x\,dx
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\end{align}\right\}\\[5pt]
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&= \frac{3}{2}\int \frac{du}{u}\\[5pt]
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&= \frac{3}{2}\ln |u|+C\\[5pt]
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&= \frac{3}{2}\ln |x^{2}+1|+C\\[5pt]
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&= \frac{3}{2}\ln (x^{2}+1) + C\,\textrm{.}
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\end{align}</math>}}
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In the last step, we take away the absolute sign around the argument in
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In the last step, we take away the absolute sign around the argument in <math>\ln</math>, because <math>x^2+1</math> is always greater than or equal to 1.
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<math>\ln </math>, because
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<math>x^{2}+1</math>
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is always greater than or equal to
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<math>\text{1}</math>.
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Version vom 14:53, 28. Okt. 2008

If we differentiate the denominator in the integrand

\displaystyle (x^2+1)' = 2x

we obtain almost the same expression as in the numerator; there is a constant 2 which is different. We therefore rewrite the numerator as

\displaystyle 3x = \frac{3}{2}\cdot 2x = \frac{3}{2}\cdot (x^2+1)',

so the integral can be written as

\displaystyle \int\frac{\tfrac{3}{2}}{x^2+1}\cdot (x^{2}+1)'\,dx\,,

and we see that the substitution \displaystyle u=x^2+1 can be used to simplify the integral,

\displaystyle \begin{align}

\int \frac{3x}{x^2+1}\,dx &= \left\{ \begin{align} u &= x^2+1\\[5pt] du &= (x^2+1)'\,dx = 2x\,dx \end{align}\right\}\\[5pt] &= \frac{3}{2}\int \frac{du}{u}\\[5pt] &= \frac{3}{2}\ln |u|+C\\[5pt] &= \frac{3}{2}\ln |x^{2}+1|+C\\[5pt] &= \frac{3}{2}\ln (x^{2}+1) + C\,\textrm{.} \end{align}

In the last step, we take away the absolute sign around the argument in \displaystyle \ln, because \displaystyle x^2+1 is always greater than or equal to 1.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:3e“