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AleksandrR [38]
3 years ago
5

Indefinite integral need help pleaseeeeeee

Mathematics
1 answer:
tatiyna3 years ago
7 0
Integrate indefinite integral:
I=\int\frac{dx}{e^{2x}+3e^x+2}dx

Solution:
1. use substitution u=e^x 
=>
du=e^xdx
=>
dx=\frac{du}{e^x}
=>
dx=\frac{du}{u}
=>
I=\int\frac{du}{u(u^2+3u+2)}du
=\int\frac{du}{u(u+2)(u+1)}du
2. decompose into partial fractions
\frac{1}{u(u+2)(u+1)}
=\frac{A}{u}+\frac{B}{u+2}+\frac{C}{u+1}
where A=1/2, B=1/2, C=-1
=\frac{1}{2u}+\frac{1}{2(u+2)}-\frac{1}{u+1}
3. Substitute partial fractions and continue
I=\int\frac{du}{2u}+\int\frac{du}{2(u+2)}-\int\frac{du}{u+1}
=\frac{log(u)}{2}+\frac{log(u+2)}{2}-log(u+1)}
4. back-substitute u=e^x
=\frac{log(e^x)}{2}+\frac{log(e^x+2)}{2}-log(e^x+1)}
=\frac{x}{2}+\frac{log(e^x+2)}{2}-log(e^x+1)}

Note: log(x) stands for natural log, and NOT log10(x)

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2 years ago
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Ainat [17]

Answer:

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Read 2 more answers
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