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

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 $$arctan(x)$$. · 4 years, 1 month ago

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

Use it and show me how do u do it... · 4 years, 1 month ago

$\frac{1}{2} e^i i \text{Ei}(-i+x)-\frac{1}{2} i e^{-i} \text{Ei}(i+x)+e^x \tan ^{-1}(x)$ · 4 years, 1 month ago

HOW??? · 4 years, 1 month ago

Wolfram | Alpha · 4 years, 1 month ago

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

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

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?) · 4 years, 1 month ago

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.. · 4 years, 1 month ago