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\int\left(\frac{\ln x-1}{(\ln x)^2+1}\right)^2 d x is equal to 

Option: 1

\frac{x}{x^2+1}+C


Option: 2

\frac{\ln x-1}{(\ln x)^2+1}


Option: 3

\frac{x}{(\ln x)^2+1}+C


Option: 4

e^x\left(\frac{x}{x^2+1}\right)+C


Answers (1)

best_answer

\begin{gathered} \int\left(\frac{\ln x-1}{(\ln x)^2+1}\right) d x . \operatorname{Let} \ln x=t \\ \\I=\int\left(\frac{t-1}{t^2+1}\right) e^t d t \end{gathered}

           \begin{aligned} I & =\int e^t\left(\frac{1}{t^2+1}+\frac{-2 t}{t^2+1}\right) d t \\ \\& =\frac{x}{(\ln x)^2+1}+C \end{aligned}

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seema garhwal

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