value of integration from e2toe4 1/x(e^((logx)^2 +1)^-1/e^((logx)2 +1)^-1 +e^((6-logx)2 +1)^-1)dx is

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āˆ« [ (1 / log x) - (1 / (log x)Ā²) ] dx āˆ« [ (1 / log x) - (1 / (log x)Ā²) ] dx
The value of integration e^2 to e^4 1/x{e^(logx)^2+1)^-1/(e^(logx)^2+1)^-1+e^(6-logx)^2+1)^-1}dx is The value of integration e^2 to e^4 1/x{e^(logx)^2+1)^-1/(e^(logx)^2+1)^-1+e^(6-logx)^2+1)^-1}dx is
Q132 | āˆ«(e^(2 logā”x )-1)/(e^(2 logā”x )+1) dx | Integral of e^(2 logā”x )-1 / e^(2 logā”x )+1 Q132 | āˆ«(e^(2 logā”x )-1)/(e^(2 logā”x )+1) dx | Integral of e^(2 logā”x )-1 / e^(2 logā”x )+1
How REAL Men Integrate Functions How REAL Men Integrate Functions
Find `I=int e^x(logx-1/(x^2))dx` Find `I=int e^x(logx-1/(x^2))dx`
`int_(e)^(e^(2)) (1/logx-1/((logx)^(2)))dx=` `int_(e)^(e^(2)) (1/logx-1/((logx)^(2)))dx=`
šŸ˜³ CLEAN BASIC CALCULUS Integrate āˆ«1/x dx=? #Shorts šŸ˜³ CLEAN BASIC CALCULUS Integrate āˆ«1/x dx=? #Shorts
integration by parts is easy integration by parts is easy