anonymous
  • anonymous
How do you find the integral of (2x^3+5x+1)e^(2x) dx ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
expand the equation first [(2x^3)e^(2x)+5x*e^(2x)+e^(2x)] dx \[\rightarrow \int\limits_{}^{}(2x^3)e ^{2x} dx +\int\limits_{}^{}5x \times e ^{2x} dx + \int\limits_{}^{} e ^{2x}dx\] to the integration by part seperately
anonymous
  • anonymous
integration by parts three times on the first integral, integrate by parts once on the second integral, and use u substitution on the third integral, then add them all together
anonymous
  • anonymous
first integral, Integrate by parts 3 times: \[\int\limits\limits 2x^3e^{2x}dx \rightarrow\] \[f(x)=e^{2x}, F(x)=\frac{e^{2x}}{2}, g(x)=x^3, g'(x)=3x^2\] \[2 ( \frac{x^3e^{2x}}{2}- \frac{3}{2}\int\limits\limits\limits x^2e^{2x}dx)\rightarrow \]\[x^3e^{2x}-3 \int\limits\limits x^2e^{2x}dx\] \[f(x)=e^{2x}, F(x)=\frac{e^{2x}}{2}, g(x)=x^2, g'(x)=2x\] \[x^3e^{2x}-3( \frac{x^2e^{2x}}{2}- \int\limits\limits\limits xe^{2x}dx)\rightarrow \]\[x^3e^{2x}-\frac{3}{2}x^2e^{2x}+3 \int\limits\limits xe^{2x}dx\] \[f(x)=e^{2x}, F(x)=\frac{e^{2x}}{2}, g(x)= x, g'(x)=1\] \[x^3e^{2x}-\frac{3}{2}x^2e^{2x}+3( \frac{xe^{2x}}{2}-\frac{1}{2} \int\limits\limits\limits e^{2x}dx) \rightarrow\] \[x^3e^{2x}-\frac{3}{2}x^2e^{2x}+\frac{3}{2}xe^{2x}-\frac{3}{2}e^{2x}+C\] second integral, integrate by parts once: \[\int\limits 5xe^{2x}dx \rightarrow\] \[f(x)=e^{2x}, F(x)=\frac{e^{2x}}{2}, g(x)=x, g'(x)=1\] \[5( \frac{xe^{2x}}{2}- \frac{1}{2} \int\limits e^{2x}dx)\rightarrow \frac{5}{2}xe^{2x}- \frac{5}{2}e^{2x}+C\] third integral, use u substitution: \[\int\limits e^{2x}dx \rightarrow \frac{1}{2}e^{2x} +C\] Your final answer should be: \[x^3e^{2x}-\frac{3}{2}x^2e^{2x}+\frac{3}{2}xe^{2x}-\frac{3}{2}e^{2x}+\frac{5}{2}xe^{2x}\] \[- \frac{5}{2}e^{2x}+\frac{1}{2}e^{2x}+C\]

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anonymous
  • anonymous
The final answer can be simplified, which I'll leave up to you
anonymous
  • anonymous
THANKS!!
anonymous
  • anonymous
no problem

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