Question

i got stuck here

Answer 1

Partial fraction

Answer 2

ugh that moment when u accidentally reload the page after typing bunch of stuffzz

Answer 3

factor the denominator \[\rm \large \frac{ 1+v }{ v^2 -3v+2}=\frac{ 1+v }{ \color{red}{(v-2)}\color{blue}{(v-1)} }\] \[ \rm \frac{ 1+v }{ \color{red}{(v-2)}\color{blue}{(v-1)} }= \frac{A}{\color{Red}{v-2}}+\frac{B}{\color{blue}{(v-1)}}\]
to rewrite the R.H.S fraction with the common denominator, multiply both sides by (v-2)(v-1)
\[ \rm (v-2)(v-1)\frac{ 1+v }{ \color{red}{(v-2)}\color{blue}{(v-1)} }= (\frac{A}{\color{Red}{v-2}}+\frac{B}{\color{blue}{v-1}}) (v-2)(v-1)\] \[ \rm \cancel{(v-2)(v-1)}\frac{ 1+v }{\cancel{ \color{red}{(v-2)}\color{blue}{(v-1)}} }= (\frac{A}{\color{Red}{v-2}}+\frac{B}{\color{blue}{v-1}}) (v-2)(v-1)\]

\[\large\rm 1+v= A(\color{blue}{v-1})+B(\color{red}{v-2})\]
there are two ways to find the value for "A" and "B"
first one is to guess a number for v that will make one term equal to 0
if we pick v=1
\[\rm 1+\color{orange}{1}=A(\color{orange}{1}-1)+B(\color{orange}{1}-2)\]\[\rm 2= \color{orange}0+B(-1)~~~~~~\rightarrow B=-2 \]
v=2 will give B(v-2) term equal to 0
\[\large\rm 1+\color{orange}{ 2}= A(\color{orange}{2}-1)+B(\color{orange}{2}-2)\]\[\rm 3=A\]
plugin A and B value and now should be easy peazy to integrate :=))
\[\int\limits_{ }^{ }\frac{ 3 }{ v-2 }+\frac{-2}{v-1} = \int\limits_{ }^{ }\frac{3}{v-2} -\int\limits_{ }^{}\frac{2}{v-1}\]

Answer 4

another method to find A and B values
\[\large\rm 1+v=Av-A+Bv-2B\]
write the coefficient of "v" term\[\large\rm\color{blue}{ 1}+(\color{orange}{1})v=(\color{orange}{A})v\color{blue}{-A}+(\color{orange}{B})v\color{blue}{-2B}\] \[\large\rm \color{orange}{1=A+B
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Answer 5

Haha, I am sorry but this is just beautiful. Not only your help @Nnesha , but that image xD

Answer 6

Lol.. Thanks :p