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gmer

  • 3 years ago

Integrate \[\int{te^{-5t} \over (5t-1)^2} dt\]

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  1. shivam_bhalla
    • 3 years ago
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    First take -5t = p -5dt = dp

  2. gmer
    • 3 years ago
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    \[-{1 \over 5} \int {te^p \over (-p -1)^2}dp\]

  3. shivam_bhalla
    • 3 years ago
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    you get \[\large \frac{1}{5^2}\int\limits_{}^{}\frac{pe^p}{(p+1)^2}dp\] \[\large \frac{1}{2*5^2}\int\limits_{}^{}\frac{(2p+2)e^p}{(p^2+1+2p)}dp - \large \frac{1}{5^2}\int\limits_{}^{}\frac{e^p}{(p+1)^2}dp\]

  4. shivam_bhalla
    • 3 years ago
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    @gmer, got it ??

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