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If dxx5x23=Ktan1f(x)+C then

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a
f(x)=53x2−1,K=15
b
f(x)=53x2−1,K=13
c
f(x)=125x2−3,K=15
d
none of these

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detailed solution

Correct option is B

Putting 5x2−3=t2 so 5xdx=tdt∫dxx5x2−3=∫xdxx25x2−3=15∫5tdtt2+3t=∫dtt2+3=13tan−1⁡t3+C=13tan−1⁡5x2−33+C.


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