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
\(\int\mathrm{e}^{tanz}z(sec^{2}z)\mathrm{d}z\).... coz there was an \(arctan(x)\) there... u have to back substitute \(arctan(x)=z\)... Now please solve this further... :-P..

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## Comments

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TopNewestHere \(tan^{-1}x\) is the inverse tangent function and can also be written as \(arctan(x)\).

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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?)

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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..

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My bad!!!...Sorry sorry...

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You should check this.

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Pls tell me how to solve this integral with the Ei(x) function..

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\[\frac{1}{2} e^i i \text{Ei}(-i+x)-\frac{1}{2} i e^{-i} \text{Ei}(i+x)+e^x \tan ^{-1}(x)\]

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HOW???

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Wolfram | Alpha

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Why not use integration by parts

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Use it and show me how do u do it...

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