Integral of (3x-2)/(x-1)^2
1 answer:
<span><span><span>(3x−2)/</span><span>(x−1<span>)^2</span></span></span>=<span>A/(<span>x−1) </span></span>+ <span><span>B/x</span><span>(x−1<span>)^2
</span></span></span> =[<span><span>A(x−1)+Bx</span><span>(x−1<span>)] / 2</span></span></span></span>
3x-2=A(x-1)+Bx
3x-2=x(A+B)-A
A+B=3
-A=-2=>A=2
A+B=3=>2+B=3=>B=1
lets check our partial fraction
we have
<span><span><span>2/(<span>x−1) </span></span>+ <span>x/<span>(x−1<span>)^2 </span></span></span>= [<span><span>2(x−1)+x] / </span><span>(x−1<span>)^2
</span></span></span> =(<span><span>3x−2) / </span><span>(x−1<span>)^2</span></span></span></span></span>
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