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Please solve this integral

\(\int \mathrm{e}^x tan^{-1}x\mathrm{d}x\)

Note by Krishna Jha
4 years, 1 month ago

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Here \(tan^{-1}x\) is the inverse tangent function and can also be written as \(arctan(x)\). Krishna Jha · 4 years, 1 month ago

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Why not use integration by parts Tanmay Bhoite · 4 years, 1 month ago

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@Tanmay Bhoite Use it and show me how do u do it... Krishna Jha · 4 years, 1 month ago

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\[\frac{1}{2} e^i i \text{Ei}(-i+x)-\frac{1}{2} i e^{-i} \text{Ei}(i+x)+e^x \tan ^{-1}(x)\] Louie Tan Yi Jie · 4 years, 1 month ago

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@Louie Tan Yi Jie HOW??? Krishna Jha · 4 years, 1 month ago

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@Krishna Jha Wolfram | Alpha Jimmi Simpson · 4 years, 1 month ago

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You should check this. Aditya Parson · 4 years, 1 month ago

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@Aditya Parson Pls tell me how to solve this integral with the Ei(x) function.. Krishna Jha · 4 years, 1 month ago

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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?) Piyal De · 4 years, 1 month ago

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@Piyal De Man... You missed out a z....the integral is \(\int\mathrm{e}^{tanz}z(sec^{2}z)\mathrm{d}z\).... coz there was an \(arctan(x)\) there... u have to back substitute \(arctan(x)=z\)... Now please solve this further... :-P.. Krishna Jha · 4 years, 1 month ago

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@Krishna Jha My bad!!!...Sorry sorry... Piyal De · 4 years, 1 month ago

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