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melmel

integral of (v-2)dv/ v^2+2v-1

  • 8 months ago
  • 8 months ago

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  1. genius12
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    \[\bf \int\limits_{}^{}\frac{ v-2 }{ v^2 + 2v-1 }dv\]Is this it? @melmel

    • 8 months ago
  2. melmel
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    \[\int\limits_{}^{} \frac{ v-2 dv }{ v ^{2}+ 2v -1}\]

    • 8 months ago
  3. melmel
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    yes

    • 8 months ago
  4. melmel
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    can you help me @genius12

    • 8 months ago
  5. melmel
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    @Callisto

    • 8 months ago
  6. genius12
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    Make the u-substitution: \(\bf u=v-2\) \(\bf \implies \frac{du}{dv}=1 \implies dv=du\) We now re-write the integral as:\[\bf \int\limits_{}^{}\frac{ u }{ (u+2)^2(2u+3) }du\]Now we use partial fractions method to solve the integral. So let's break this in to two fractions:\[\bf \int\limits_{}^{}\frac{ Au+B }{ (u+2)^2 }+\frac{ C }{ 2u+3 }du\]Now we solve for \(\bf A, B, C\):\[\bf (Au+B)(2u+3)+C(u+2)^2=u\]By setting \(\bf u=-\frac{3}{2}\), we get:\[\bf \frac{ C }{ 4 }=-\frac{ 3 }{ 2 } \implies C = -6\]Now by setting \(\bf u=0\), we get:\[\bf 3B=24 \implies B =8\]Now let's choose \(\bf u = 1\), which gives:\[\bf (A+8)(5)=55 \implies A+8 =11 \implies A = 3\]Now we re-write the integral as:\[\bf \int\limits_{}^{}\frac{ 3u+8 }{ (u+2)^2}+\frac{ -6 }{ 2u+3 }du=\int\limits_{}^{}\frac{ 3u+8 }{ (u+2)^2}du-\int\limits_{}^{}\frac{ 6 }{ 2u+3 }du\]

    • 8 months ago
  7. genius12
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    @melmel Do you understand what I did and can you carry on? Or do you still need me to continue further?

    • 8 months ago
  8. melmel
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    continue genius12

    • 8 months ago
  9. genius12
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    As you can see, the right integral with integrand 6/2u+3 can be evaluated easily but the left integral is a bit more difficult. To evaluate the left integral, we make another substitution: \[\bf m=u+2 \implies dm=du\]This also implies that:\[\bf 3u+8=3m+2\]Now we re-write the left more complicated integral as:\[\bf \int\limits_{}^{}\frac{ 3m+2 }{ m^2 } dm=\int\limits_{}^{} 3m^{-1}+2 m^{-2} dm=3\int\limits_{}^{}\frac{1}{m}dm+2\int\limits_{}^{}m^{-2} dm\]\[\bf = 3\ln|m|+(2(-m^{-1}))=3\ln|m|-2m^{-1}\]Now that we have evaluated the left integral, we now evaluate the right integral which is pretty easy:\[\bf \int\limits_{}^{}\frac{6}{2u+3}du=6\int\limits_{}^{}\frac{1}{2u+3}du=3\ln|2u+3|\]

    • 8 months ago
  10. cambrige
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    Isnt there any easier method genius12??

    • 8 months ago
  11. genius12
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    Now we substitute everything back in to get everything in terms of 'v' again. Hence:\[\bf 3\ln|m|-2m^{-1}=3\ln|u+2|-2(u+2)^{-1}=3\ln|v|-2v^{-1}\]\[\bf 3\ln|2u+3|=3\ln|2(v-2)+3|=3\ln|2v+1|\]Now to put it all together we get:\[\bf \int\limits_{}^{}\frac{v-2}{v^2+2v-1}dv=3\ln|v|-2v^{-1}-3\ln|2v+1|+C\]

    • 8 months ago
  12. litchlani
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    i dont think its even correct ? how u get to that integral with u=v-2 substitution u had (u+2)^2(2u+3) if we substitude back we get (v-2+2)^2(2(v-2)+3)= v^2(2v-4+3)=v^2(2v-1)=2v^3-v^2 doesnt seem same to what u had before substitution

    • 8 months ago
  13. genius12
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    i might've made a mistake somewhere since its 10:50 am and i havent even slept and i am fasting...

    • 8 months ago
  14. litchlani
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    since first step is wrong , its all wrong ..

    • 8 months ago
  15. genius12
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    @litchlani If you had to solve this integral, it would work in the same way. @melmel Now knows the approach of how to do it. Making a numerical mistake isn't what's important here. When you know how to do something, you can do it yourself even if there is a mistake.

    • 8 months ago
  16. litchlani
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    yeah true that , you would need partial fraction still , you just dont need that substitution at the begining , you probably should rewrite the bottom equation as (x+(1+2^(1/2)))(x+(1-2^(1/2))) then look for partial fractions and then u can solve the integral

    • 8 months ago
  17. genius12
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    @litchlani I notice the dumb mistake I made out of sleepiness lol. When I made the first u-substitution, I multiplied the stuff in the denominator instead of adding -.-

    • 8 months ago
  18. genius12
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    @melmel I made some wrong steps out of sleepiness especially my first step where I made the substitution, I incorrectly multiplied the factors in the denominator instead of adding lol. So don't follow the solution but still look at the steps because now you can figure out how to solve it yourself.

    • 8 months ago
  19. genius12
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    I ended up solving for the integral of (7v+2)/(2v^3+v^2) because of accidentally multiplying instead of adding when I made the first substitution. Regardless, never feel sad, stupid mistakes happen when your eyes are about to collapse. You should however like I mentioned already, take a look at the steps still judge how you can solve the integral yourself through similar steps because the idea will be the same.

    • 8 months ago
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