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Integrate \int \frac{dx}{(x+1)(x-2)(x-1)}

  • Option 1)

    1/6 ln (x^{2}-1)(x-2)+C

  • Option 2)

    1/6 ln (x^{2}-1)/(x-2)+C

  • Option 3)

    1/6 ln (x^{2}-1)(x-2)^{2}+C

  • Option 4)

    1/6 ln (x^{2}-1)/(x-2)^{2}+C

 

Answers (1)

best_answer

As we have learned

Rule of integration by Partial fraction -

Linear and non-repeated:

\frac{P(x)}{Q(x)}=\frac{P(x)}{(x-\alpha _{1})(x-\alpha _{2})\cdot \cdot \cdot (x-\alpha _{n})}

Let  \frac{P(x)}{Q(x)}=\frac{A}{(x-\alpha _{1})}+\frac{B}{(x-\alpha _{2})}\cdot \cdot \cdot

Find A,B...

By comparing N^{x} and  P(x) 

-

 

 \int \frac{dx}{(x+1)(x-2)(x-1)}= \int \frac{A}{(x+1)}+\frac{B}{(x-2)}+\frac{C}{x-1}dx

Thus 1= A(x-2)(x-1)+ B (x+1)(x-1)+C(x+1)(x-2)

put x= 1

\Rightarrow c= 1/6

put x= +2

\Rightarrow B= 1/3

put x= -1 

\Rightarrow A= 1/6

thus I = 1/6 ln (x^{2}-1)+ 1/3 ln (x-2)+ C

I = 1/6 ln (x^{2}-1)(x-2)^{2}+ C

 

 

 

 

 


Option 1)

1/6 ln (x^{2}-1)(x-2)+C

This is incorrect

Option 2)

1/6 ln (x^{2}-1)/(x-2)+C

This is incorrect

Option 3)

1/6 ln (x^{2}-1)(x-2)^{2}+C

This is correct

Option 4)

1/6 ln (x^{2}-1)/(x-2)^{2}+C

This is incorrect

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Aadil

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