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a
f(x)=tan−1x,K=−1
b
f(x)=tan−1x,K=1
c
f(x)=2tan−1x,K=−1
d
f(x)=2tan−1x,K=1
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Correct option is A
∫x1+x2dx=12∫dtt=12⋅t1/2=t=1+x2 [where 1 + x2 = t]we integrate by the parts, taking tan−1x as the first function and x/1+x2 as the second.⇒∫xtan−1x1+x2dx=tan−1x1+x2−∫11+x2⋅1+x2dx=1+x2tan−1x−∫dx1+x2=1+x2tan−1x−logx+x2+1+CThus f(x)=tan−1x,K=−1.Similar Questions
is equal to
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