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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) |
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Answer» 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) |
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