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Take z=arctan(x), then take dz =dx/(1+x^2)=dx/(1+tan^2z)=dx/(sec^2x)..................so dx=dz(sec^2z)...............so the integral reduces to [ integral e^(tan(z)) sec^2(z)) dz.....now substitute y=tan(z).....and tada u'll ger it!! Hope u understand it.........
(N.B..here sec^2(z) means sec squared z.................okay?)
Man... You missed out a z....the integral is
∫etanzz(sec2z)dz.... coz there was an arctan(x) there... u have to back substitute arctan(x)=z... Now please solve this further... :-P..
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Here tan−1x is the inverse tangent function and can also be written as arctan(x).
Take z=arctan(x), then take dz =dx/(1+x^2)=dx/(1+tan^2z)=dx/(sec^2x)..................so dx=dz(sec^2z)...............so the integral reduces to [ integral e^(tan(z)) sec^2(z)) dz.....now substitute y=tan(z).....and tada u'll ger it!! Hope u understand it.........
(N.B..here sec^2(z) means sec squared z.................okay?)
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Man... You missed out a z....the integral is ∫etanzz(sec2z)dz.... coz there was an arctan(x) there... u have to back substitute arctan(x)=z... Now please solve this further... :-P..
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My bad!!!...Sorry sorry...
You should check this.
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Pls tell me how to solve this integral with the Ei(x) function..
21eiiEi(−i+x)−21ie−iEi(i+x)+extan−1(x)
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HOW???
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Wolfram | Alpha
Why not use integration by parts
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Use it and show me how do u do it...