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Question

If dx(x2+1)(x2+4)=K tan1x+L tan1x2+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 A 16
1(x2+1)(x2+4)=13.1(x2+1)13.1(x2+4)
dx(x2+1)(x2+4)=13tan1x16tan1(x2)+C
Comparing with R.H.S., we have k=13,L=16
K+L=16

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