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kann mir jemand bitte erklären wie man dieses Integral löst?

\(\int_{-3}^{2} \frac{7x}{1+x^2}\)

 

die lösung wäre \(7ln(x^2+1) / 2\)

 19.11.2015

Beste Antwort 

 #1
avatar+26270 
+15

kann mir jemand bitte erklären wie man dieses Integral löst?

\( \int \limits_{-3}^{2} { \frac{7x}{1+x^2}\ dx}\)

 

\(\small{ \begin{array}{lrcll} & \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} &=& 7 \int \limits_{-3}^{2} { \dfrac{x}{1+x^2}\ dx} \\ \hline \text{Substitution 1: } & x &=& \sinh{(z)} \\ & \ dx &=& \cosh{(z)} \ dz\\ \cosh^2{(z)}- \sinh^2{(z)} = 1 \\ \cosh^2{(z)} = 1 + \sinh^2{(z)} \\ & 1+x^2 &=& 1 + \sinh^2{(z)} =\cosh^2{(z)} \\ \hline & 7 \int \limits_{-3}^{2} { \dfrac{x}{1+x^2}\ dx} &=& 7 \int \limits_{-3}^{2} { \dfrac{\sinh{(z)}}{\cosh^2{(z)}} \cdot \cosh{(z)} \ dz}\\ & &=& 7 \int \limits_{-3}^{2} { \dfrac{\sinh{(z)}}{\cosh{(z)}} \ dz}\\ \sinh{(z)} = \frac{ e^z-e^{-z} }{2} \\ \cosh{(z)} = \frac{ e^z+e^{-z} }{2} \\ & &=& 7 \int \limits_{-3}^{2} { \dfrac{e^z-e^{-z}}{e^z+e^{-z}} \ dz}\\ \hline \text{Substitution 2: } & u &=& e^z+e^{-z} \\ & \ du &=& e^z\ dz+(-1)e^{-z} \ dz\\ & \ du &=& (e^z-e^{-z}) \ dz\\ \hline = 7 \int \limits_{-3}^{2} { \dfrac{\ du}{u}}\\ = 7 [\ln{(u)}]_{-3}^{2} & | \quad u = e^z+e^{-z}\\ = 7 [\ln{(e^z+e^{-z})}]_{-3}^{2} & | \quad e^z+e^{-z} = 2\cosh{(z)}\\ = 7 [\ln{(~2\cosh{(z)}~)}]_{-3}^{2} & | \quad \cosh^2{(z)} = 1 + \sinh^2{(z)} \\ = 7 [\ln{(~2 \sqrt{1 + \sinh^2{(z)}} ~)}]_{-3}^{2} & | \quad \sinh{(z)} = x \\ = 7 [\ln{(~2 \sqrt{1 + x^2} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{(~2 \sqrt{1 + x^2} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{(~2 (1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{2}+\ln{(~2 (1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\underbrace{\ln{2}}_{\ln{2}-\ln{2}\ \text{kürzt sich raus}}+\ln{(~(1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{(~ (1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\frac12 \ln{(~ 1 + x^2 ~)}]_{-3}^{2} \\ & &=& \frac72 [ \ln{(~ 1 + x^2 ~)}]_{-3}^{2} \\ & \mathbf{ \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} }& \mathbf{=}& \mathbf{ \frac72 [ \ln{(~ 1 + x^2 ~)}]_{-3}^{2} } \\ \hline & \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} &=& \frac72 [ \ln{(~ 1 + x^2 ~)}]_{-3}^{2} \\ & &=& \frac72 [ \ln{(~ 1 + (2)^2 ~)} - \ln{(~ 1 + (-3)^2 ~)} ] \\ & &=& \frac72 [ \ln{(~ 1 + 4 ~)} - \ln{(~ 1 + 9 ~)} ] \\ & &=& \frac72 [ \ln{(~ 5 ~)} - \ln{(~ 10 ~)} ] \\ & &=& \frac72 \ln{( \frac{5}{10} )} \\ & &=& \frac72 \ln{( \frac12 )} \\ & &=& \frac72 [ \ln{( 1 )} - \ln{( 2 )} ] \qquad \ln{( 1 )} = 0\\ & &=& \frac72 [ 0 - \ln{( 2 )} ] \\ & &=& \frac72 [ -\ln{( 2 )} ] \\ & &=& -\frac72 \ln{( 2 )} \\ & \mathbf{ \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} }&\mathbf{=}& \mathbf{ -2.4260151320 } \\ \end{array} }\)

 

laugh

 20.11.2015
bearbeitet von heureka  20.11.2015
bearbeitet von heureka  20.11.2015
bearbeitet von heureka  20.11.2015
bearbeitet von heureka  20.11.2015
 #1
avatar+26270 
+15
Beste Antwort

kann mir jemand bitte erklären wie man dieses Integral löst?

\( \int \limits_{-3}^{2} { \frac{7x}{1+x^2}\ dx}\)

 

\(\small{ \begin{array}{lrcll} & \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} &=& 7 \int \limits_{-3}^{2} { \dfrac{x}{1+x^2}\ dx} \\ \hline \text{Substitution 1: } & x &=& \sinh{(z)} \\ & \ dx &=& \cosh{(z)} \ dz\\ \cosh^2{(z)}- \sinh^2{(z)} = 1 \\ \cosh^2{(z)} = 1 + \sinh^2{(z)} \\ & 1+x^2 &=& 1 + \sinh^2{(z)} =\cosh^2{(z)} \\ \hline & 7 \int \limits_{-3}^{2} { \dfrac{x}{1+x^2}\ dx} &=& 7 \int \limits_{-3}^{2} { \dfrac{\sinh{(z)}}{\cosh^2{(z)}} \cdot \cosh{(z)} \ dz}\\ & &=& 7 \int \limits_{-3}^{2} { \dfrac{\sinh{(z)}}{\cosh{(z)}} \ dz}\\ \sinh{(z)} = \frac{ e^z-e^{-z} }{2} \\ \cosh{(z)} = \frac{ e^z+e^{-z} }{2} \\ & &=& 7 \int \limits_{-3}^{2} { \dfrac{e^z-e^{-z}}{e^z+e^{-z}} \ dz}\\ \hline \text{Substitution 2: } & u &=& e^z+e^{-z} \\ & \ du &=& e^z\ dz+(-1)e^{-z} \ dz\\ & \ du &=& (e^z-e^{-z}) \ dz\\ \hline = 7 \int \limits_{-3}^{2} { \dfrac{\ du}{u}}\\ = 7 [\ln{(u)}]_{-3}^{2} & | \quad u = e^z+e^{-z}\\ = 7 [\ln{(e^z+e^{-z})}]_{-3}^{2} & | \quad e^z+e^{-z} = 2\cosh{(z)}\\ = 7 [\ln{(~2\cosh{(z)}~)}]_{-3}^{2} & | \quad \cosh^2{(z)} = 1 + \sinh^2{(z)} \\ = 7 [\ln{(~2 \sqrt{1 + \sinh^2{(z)}} ~)}]_{-3}^{2} & | \quad \sinh{(z)} = x \\ = 7 [\ln{(~2 \sqrt{1 + x^2} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{(~2 \sqrt{1 + x^2} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{(~2 (1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{2}+\ln{(~2 (1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\underbrace{\ln{2}}_{\ln{2}-\ln{2}\ \text{kürzt sich raus}}+\ln{(~(1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\ln{(~ (1 + x^2)^{\frac12} ~)}]_{-3}^{2} \\ & &=& 7 [\frac12 \ln{(~ 1 + x^2 ~)}]_{-3}^{2} \\ & &=& \frac72 [ \ln{(~ 1 + x^2 ~)}]_{-3}^{2} \\ & \mathbf{ \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} }& \mathbf{=}& \mathbf{ \frac72 [ \ln{(~ 1 + x^2 ~)}]_{-3}^{2} } \\ \hline & \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} &=& \frac72 [ \ln{(~ 1 + x^2 ~)}]_{-3}^{2} \\ & &=& \frac72 [ \ln{(~ 1 + (2)^2 ~)} - \ln{(~ 1 + (-3)^2 ~)} ] \\ & &=& \frac72 [ \ln{(~ 1 + 4 ~)} - \ln{(~ 1 + 9 ~)} ] \\ & &=& \frac72 [ \ln{(~ 5 ~)} - \ln{(~ 10 ~)} ] \\ & &=& \frac72 \ln{( \frac{5}{10} )} \\ & &=& \frac72 \ln{( \frac12 )} \\ & &=& \frac72 [ \ln{( 1 )} - \ln{( 2 )} ] \qquad \ln{( 1 )} = 0\\ & &=& \frac72 [ 0 - \ln{( 2 )} ] \\ & &=& \frac72 [ -\ln{( 2 )} ] \\ & &=& -\frac72 \ln{( 2 )} \\ & \mathbf{ \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} }&\mathbf{=}& \mathbf{ -2.4260151320 } \\ \end{array} }\)

 

laugh

heureka 20.11.2015
bearbeitet von heureka  20.11.2015
bearbeitet von heureka  20.11.2015
bearbeitet von heureka  20.11.2015
bearbeitet von heureka  20.11.2015
 #2
avatar+26270 
+15

kann mir jemand bitte erklären wie man dieses Integral löst?

 

2. Möglichkeit - einfacher

 

\(\small{ \begin{array}{lrcll} & \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} &=& 7 \int \limits_{-3}^{2} { \dfrac{x}{1+x^2}\ dx} \\ \hline \text{Substitution : } & u &=& 1+x^2 \ dx\\ & \ du &=& 2x \ dx \qquad \ dx = \frac{\ du}{2x} \\ \hline & 7 \int \limits_{-3}^{2} { \dfrac{x}{1+x^2}\ dx} &=& 7 \int \limits_{-3}^{2} { \dfrac{x}{u} \cdot \dfrac{\ du}{2x} }\\ & &=& \frac72 \int \limits_{-3}^{2} { \dfrac{\ du}{u} }\\ \boxed{~ \text{Formel: } \int \limits { \dfrac{\ du}{u}} = \ln{(u)} + c ~}\\ & &=& \frac72 [~ \ln{(u)} ~]_{-3}^{2} \qquad | u = 1+x^2 \\ & \mathbf{ \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} } & \mathbf{=} & \mathbf{ \frac72 [~ \ln{(1+x^2 )} ~]_{-3}^{2} }\\ \hline & \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} &=& \frac72 [~ \ln{(~ 1 + x^2 ~)} ~]_{-3}^{2} \\ & &=& \frac72 [~ \ln{(~ 1 + (2)^2 ~)} - \ln{(~ 1 + (-3)^2 ~)} ~] \\ & &=& \frac72 [~ \ln{(~ 1 + 4 ~)} - \ln{(~ 1 + 9 ~)} ~] \\ & &=& \frac72 [~ \ln{(~ 5 ~)} - \ln{(~ 10 ~)} ~] \\ & &=& \frac72 \ln{( \frac{5}{10} )} \\ & &=& \frac72 \ln{( \frac12 )} \\ & &=& \frac72 [ \ln{( 1 )} - \ln{( 2 )} ] \qquad \ln{( 1 )} = 0\\ & &=& \frac72 [ 0 - \ln{( 2 )} ] \\ & &=& \frac72 [ -\ln{( 2 )} ] \\ & &=& -\frac72 \ln{( 2 )} \\ & \mathbf{ \int \limits_{-3}^{2} { \dfrac{7x}{1+x^2}\ dx} }&\mathbf{=}& \mathbf{ -2.4260151320 } \\ \end{array} }\)

laugh

 20.11.2015
bearbeitet von heureka  20.11.2015
 #3
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+5

Vielen vielen herzlichen dank !!! smiley

 20.11.2015

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