Math, asked by Kriti0184, 8 months ago

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Answered by susam3094
2

3x+5/x^3-x^2-x+1

--->3x+5/x^2(x-1)-(x-1)

=3x+5/(x-1)(x^2-1)

=3x+5/(x-1)^2(x+1)

Let 3x+5/(x-1)^2(x+1)=A/(x-1) +B/(x-1)^2 +C/x+1

Then,

3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^2......(i)

Putting x=1,-1 and 0 successively in (i),we get

8=2B,2=4C and 5= -A+B+C

--->B=4,C=1/2,A=-1/2

Therefore,

3x+5/(x-1)^2(x+1)=-1/2(x-1) +4/(x-1)^2 +1/2(x+1)

--->integration of 3x+5/(x-1)^2(x+1) =-1/2 log|x-1| -4/x-1 +1/2log|x+1|+ c

=1/2log|x+1/x-1| -4/x-1 +c

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Answered by ramlalrastogi8
1

Answer:See below

Step-by-step explanation:

3x+5/x^3-x^2-x+1

--->3x+5/x^2(x-1)-(x-1)

=3x+5/(x-1)(x^2-1)

=3x+5/(x-1)^2(x+1)

Let 3x+5/(x-1)^2(x+1)=A/(x-1) +B/(x-1)^2 +C/x+1

Then,

3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^2......(i)

Putting x=1,-1 and 0 successively in (i),we get

8=2B,2=4C and 5= -A+B+C

--->B=4,C=1/2,A=-1/2

Therefore,

3x+5/(x-1)^2(x+1)=-1/2(x-1) +4/(x-1)^2 +1/2(x+1)

--->integration of 3x+5/(x-1)^2(x+1) =-1/2 log|x-1| -4/x-1 +1/2log|x+1|+ c

=1/2log|x+1/x-1| -4/x-1 +c

mark as brainliest

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