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Most challenging one!!!
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| Forum Index -> Integral Calculus -> View Full Question |
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Check my method Shakir Put 1+x4=x4y4 x4=1/(y4-1) dx= -y3dy/x3(y4-1)2 So given integral is  -y3dy/[x4y(y4-1)2]= -y2dy/x4(y4-1)2 = - y2dy/(y4-1)So final answer is (1/2)[(1/2)log{(1+y)/(1-y)]-tan-1y}+c So which one you think is shorter???  |
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-ADARSH NITK Surathkal |
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