If ∫dx(x2+1)(x2+4)=Ktan−1x+Ltan−1x2+c where c is arbitrary constant, then K + L =
A
16
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B
13
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C
23
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D
0
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Solution
The correct option is A16 1(x2+1)(x2+4)=13.1(x2+1)−13.1(x2+4) ∴∫dx(x2+1)(x2+4)=13tan−1x−16tan−1(x2)+C Comparing with R.H.S., we have k=13,L=−16 ∴K+L=16