Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message

jsMath

Lösung 2.2:3e

Aus Online Mathematik Brückenkurs 2

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

Version vom 14:53, 28. Okt. 2008

If we differentiate the denominator in the integrand

(x2+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

3x=232x=23(x2+1)

so the integral can be written as

23x2+1(x2+1)dx 

and we see that the substitution u=x2+1 can be used to simplify the integral,

3xx2+1dx=udu=x2+1=(x2+1)dx=2xdx=23udu=23lnu+C=23lnx2+1+C=23ln(x2+1)+C.

In the last step, we take away the absolute sign around the argument in ln, because x2+1 is always greater than or equal to 1.

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