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Question Number 200048 by Calculusboy last updated on 12/Nov/23
Commented by 0670322918 last updated on 13/Nov/23
∫tan−1(x)∫tan−1(x)dxdx=f(x)=∫tan−1(x)dx=xtan−1(x)−12ln(x2+1)+cf′(x)=tan−1(x)∫f′(x)f(x)dx=ln∣f(x)∣+c∫tan−1(x)∫tan−1(x)dxdx=ln∣xtan−1(x)−12ln(x2+1)+c∣+c1
Commented by Calculusboy last updated on 13/Nov/23
thankssir
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