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calculate the integral \int \frac{x^{3}+10x^{2}+3x+36}{(x-1)(x^{2}+4)^{2}}dx

  • Option 1)

    2ln |x-1|+ln |x^{2}-4|+1/2 \tan ^{-1}x/2+1/2 (x^{2}+4)+C

  • Option 2)

    2ln |x-1|-ln |x^{2}-4|+1/2 \tan ^{-1}x/2+1/2 (x^{2}+4)+C

  • Option 3)

    2ln |x-1|-ln |x^{2}-4|-1/2 \tan ^{-1}x/2+1/2 (x^{2}+4)+C

  • Option 4)

    2ln |x-1|-ln |x^{2}-4|-1/2 \tan ^{-1}x/2-1/2 (x^{2}+4)+C

 

Answers (1)

best_answer

As we have learned

Rule for Partial fraction -

Quadratic Repeated :

Q(x)=(x-\alpha _{1})(x-\alpha _{2})\cdot \cdot \cdot (ax^{2}+bx+c)^{2}

\therefore \frac{Ax+B}{(ax^{2}+bx+c)}+\frac{Cx+D}{(ax^{2}+bx+c)^{2}}+\frac{E}{(x-\alpha _{1})}+\frac{F}{(x-\alpha _{2})}+\cdot \cdot \cdot

- wherein

find A,B,C,D,E by comparing P(x)

 

 \frac{x^{3}+10x^{2}+3x+36}{(x-1)(x^{2}+4)^{2}}= \frac{A}{x-1}+\frac{Bx+C}{x^{2}+4}+\frac{Dx+E}{(x^{2}+4)^{2}}

On solving 

 

 A=2;  B= -2  ; C=-1 ; D= 1 ; E = 0

Thus I = 2ln |x-1|-ln |x^{2}+4|-1/2 \tan ^{-1} x/2 - 1/2 (x^{2}+4)+ C

 

 

 


Option 1)

2ln |x-1|+ln |x^{2}-4|+1/2 \tan ^{-1}x/2+1/2 (x^{2}+4)+C

This is incorrect

Option 2)

2ln |x-1|-ln |x^{2}-4|+1/2 \tan ^{-1}x/2+1/2 (x^{2}+4)+C

This is incorrect

Option 3)

2ln |x-1|-ln |x^{2}-4|-1/2 \tan ^{-1}x/2+1/2 (x^{2}+4)+C

This is incorrect

Option 4)

2ln |x-1|-ln |x^{2}-4|-1/2 \tan ^{-1}x/2-1/2 (x^{2}+4)+C

This is correct

Posted by

gaurav

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