Kann mir ein Moderator bitte helfen!
\(\int_{1}^{3} \frac{x-4}{2x+1} *dx\)
+26367 \(\begin{array}{rcll} \int \limits_{1}^{3} \frac{x-4}{2x+1} \ dx = \ ? \end{array}\)
\(\begin{array}{rcll} \int \limits_{1}^{3} \frac{x-4}{2x+1} \ dx &= & \int \limits_{1}^{3} \frac{x}{2x+1} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \int \limits_{1}^{3} \frac{x}{2(x+\frac12)} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \frac{x}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \frac{x+\frac12-\frac12}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \ dx +\frac12 \int \limits_{1}^{3} \frac{-\frac12}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \ dx -\frac14 \int \limits_{1}^{3} \frac{1}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \ dx -\frac94 \int \limits_{1}^{3} \frac{1}{x+\frac12} \ dx \\ &= & \frac12 [x]_1^3 -\frac94 [\ln{(x+\frac12)}]_1^3 \\ &= & \frac12 (3-1) -\frac94 [\ln{(3+\frac12)}-\ln{(1+\frac12)} ] \\ &= & \frac12 (3-1) -\frac94 [\ln{ (\frac72)}-\ln{(\frac32)} ] \\ &= & \frac12 (3-1) -\frac94 [\ln{ \left( \frac{ \frac72 }{ \frac32 } \right) } ] \\ &= & \frac12 (3-1) -\frac94 [\ln{ ( \frac{ 7 }{ 3 } ) } ] \\ &= & \frac12 (2) -\frac94 [\ln{ ( \frac{ 7 }{ 3 } ) } ] \\ &= & 1 -\frac94 [\ln{ ( \frac{ 7 }{ 3 } ) } ] \\ &=& 1-2,25\cdot 0,84729786039\\ &=& 1- 1,90642018587\\ &=& -0,90642018587\\ \mathbf{ \int \limits_{1}^{3} \frac{x-4}{2x+1} \ dx } &\mathbf{=}& \mathbf{-0,90642018587} \end{array}\)
+26367 \(\begin{array}{rcll} \int \limits_{1}^{3} \frac{x-4}{2x+1} \ dx = \ ? \end{array}\)
\(\begin{array}{rcll} \int \limits_{1}^{3} \frac{x-4}{2x+1} \ dx &= & \int \limits_{1}^{3} \frac{x}{2x+1} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \int \limits_{1}^{3} \frac{x}{2(x+\frac12)} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \frac{x}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \frac{x+\frac12-\frac12}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \ dx +\frac12 \int \limits_{1}^{3} \frac{-\frac12}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \ dx -\frac14 \int \limits_{1}^{3} \frac{1}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \ dx -\frac94 \int \limits_{1}^{3} \frac{1}{x+\frac12} \ dx \\ &= & \frac12 [x]_1^3 -\frac94 [\ln{(x+\frac12)}]_1^3 \\ &= & \frac12 (3-1) -\frac94 [\ln{(3+\frac12)}-\ln{(1+\frac12)} ] \\ &= & \frac12 (3-1) -\frac94 [\ln{ (\frac72)}-\ln{(\frac32)} ] \\ &= & \frac12 (3-1) -\frac94 [\ln{ \left( \frac{ \frac72 }{ \frac32 } \right) } ] \\ &= & \frac12 (3-1) -\frac94 [\ln{ ( \frac{ 7 }{ 3 } ) } ] \\ &= & \frac12 (2) -\frac94 [\ln{ ( \frac{ 7 }{ 3 } ) } ] \\ &= & 1 -\frac94 [\ln{ ( \frac{ 7 }{ 3 } ) } ] \\ &=& 1-2,25\cdot 0,84729786039\\ &=& 1- 1,90642018587\\ &=& -0,90642018587\\ \mathbf{ \int \limits_{1}^{3} \frac{x-4}{2x+1} \ dx } &\mathbf{=}& \mathbf{-0,90642018587} \end{array}\)
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