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If ∫x2−1x4+x2+1dx=12ln∣∣∣x2−x+Kx2+x+K∣∣∣+C, Then value of K is (where K is fixed constant and C is integration constant)

Answer» If x21x4+x2+1dx=12lnx2x+Kx2+x+K+C, Then value of K is (where K is fixed constant and C is integration constant)


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