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PLease solve it for me
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| Forum Index -> Integral Calculus -> View Full Question |
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 dx/(1+x3)=dx/[(1+x)(1-x+x2)]Let 1/[(1+x)(1-x+x2)]=A/(1+x)+(Bx+C)/(1-x+x2) Find A,B and C we get A=1/3 ,B= -1/3 and C=2/3 So dx/(1+x3)=[1/3(1+x)+(2-x)/3(1-x+x2)]dxThen you can proceed. The final answer is [ln(1+x)]/3-[log(1-x+x2)]/6+{tan-1[(2x-1)/
 3]}/3 |
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-ADARSH NITK Surathkal |
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