Question

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

Answer 1

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

Answer 2

\[\int\limits_{}^{} \frac{ v-2 dv }{ v ^{2}+ 2v -1}\]

Answer 3

yes

Answer 4

can you help me @genius12

Answer 5

@Callisto

Answer 6

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:\[\bf \int\limits_{}^{}\frac{ 3u+8 }{ (u+2)^2}+\frac{ -6 }{ 2u+3 }du=\int\limits_{}^{}\frac{ 3u+8 }{ (u+2)^2}du-\int\limits_{}^{}\frac{ 6 }{ 2u+3 }du\]

Answer 7

@melmel Do you understand what I did and can you carry on? Or do you still need me to continue further?

Answer 8

continue genius12

Answer 9

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:\[\bf \int\limits_{}^{}\frac{6}{2u+3}du=6\int\limits_{}^{}\frac{1}{2u+3}du=3\ln|2u+3|\]

Answer 10

Isnt there any easier method genius12??

Answer 11

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:\[\bf \int\limits_{}^{}\frac{v-2}{v^2+2v-1}dv=3\ln|v|-2v^{-1}-3\ln|2v+1|+C\]

Answer 12

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

Answer 13

i might've made a mistake somewhere since its 10:50 am and i havent even slept and i am fasting...

Answer 14

since first step is wrong , its all wrong ..

Answer 15

@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.

Answer 16

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

Answer 17

@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 -.-

Answer 18

@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.

Answer 19

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.