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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: cbse sample paper doubt
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[Post New]19 Mar 2008 15:40:13 IST
    Subject: cbse sample paper doubt
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jessforyou (2)

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integrate : [(x2+1)(x2+4)]dx / [(x2+3)(x2-5)]
    
[Post New]19 Mar 2008 15:43:14 IST
    Subject: Re:cbse sample paper doubt
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layman (148)

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First divide and get it as,
=1+(7x^2 +19)/[(x2+3)(x2-5)]
Then do partial fractions and hence get the answer


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[Post New]19 Mar 2008 16:05:08 IST
    Subject: Re:cbse sample paper doubt
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ramkumar_november (1270)

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we can also do partial fraction directly.......
 
\frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}\;\;=\;\;1+\frac{A}{x^2+3}+\frac{B}{x^2-5}
 
substitute values for x and find A and B  ......
 
 
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[Post New]19 Mar 2008 16:06:08 IST
    Subject: Re:cbse sample paper doubt
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layman (148)

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Yes that's what I meant


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[Post New]19 Mar 2008 16:22:41 IST
    Subject: Re:cbse sample paper doubt
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sboosy (3063)

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[tex] \\ \int \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}\,dx \\ \\ \int \frac{(x^4+5x^2+4)}{(x^4-2x^2-15)}\,dx \\ \\ (x^4+5x^2+4) = (x^4-2x^2-15) + (7x^2+19) \\ \\ \int [1+\frac{(7x^2+19)}{(x^2+3)(x^2-5)}]\,dx \\ \\ x+ \int \frac{8(x^2+3) -1(x^2-5) - 10}{(x^2+3)(x^2-5)}\,dx \\ \\ x+ 8 \int \frac{1}{x^2-5}\,dx -1 \int \frac{1}{x^2+3}\,dx - \frac{10}{8} \int \frac{(x^2+3)-(x^2-5)}{(x^2+3)(x^2-5)}\,dx \\ \\ x+\frac{27}{4}\int \frac{1}{x^2-5}\,dx + \frac{1}{4}\int \frac{1}{x^2+3}\,dx + C \\ \\ \mbox{Now both the integrals are in standard form}[\tex]
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