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sjg13e

  • one year ago

Help with a calculus problem (integration by parts)

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  1. sjg13e
    • one year ago
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    \[integrate \ln(2x+1)dx\]

  2. sjg13e
    • one year ago
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    I have some work, give me a few minutes to input it

  3. sjg13e
    • one year ago
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    \[\int\limits_{}^{}\ln(2x+1)dx = x*\ln(2x+1) - \int\limits_{}^{}\frac{ 2x }{ 2x +1 }\]

  4. sjg13e
    • one year ago
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    I don't know how to take the integral of 2x / 2x + 1. I've viewed solutions online and I've seen...

  5. sjg13e
    • one year ago
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  6. sjg13e
    • one year ago
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    I don't know why they split it into (2x - 1 + 1)/2x +1 Everything else I can do, except I just don't understand that part

  7. ganeshie8
    • one year ago
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    do you still remember how to do "long division" ?

  8. ganeshie8
    • one year ago
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    |dw:1440095909160:dw|

  9. sjg13e
    • one year ago
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    lol, i'm not gonna lie, it's been a while. but i can do it! thanks for answering. The way I initially approached it was by doing \[2\int\limits_{}^{}x * \frac{ 1 }{ 2x +1 }\] and then using \[uv - \int\limits_{}^{}vdu\] to solve. so maybe that's why i didn't initially think of doing long division

  10. ganeshie8
    • one year ago
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    that gives you back the original integrand haha!

  11. sjg13e
    • one year ago
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    yeah that's why i was like what am i doing wrong?? haha

  12. ganeshie8
    • one year ago
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    ikr! review this quick for long division https://www.khanacademy.org/math/algebra2/polynomial_and_rational/dividing_polynomials/v/polynomial-division

  13. sjg13e
    • one year ago
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    thanks so much for your help! i'll check the link out

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