Please solve this integral $\int \mathrm{e}^x tan^{-1}x\mathrm{d}x$ Note by Krishna Jha
6 years, 3 months ago

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

- 6 years, 3 months 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?)

- 6 years, 3 months 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..

- 6 years, 3 months ago

- 6 years, 3 months ago

You should check this.

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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

- 6 years, 3 months ago

HOW???

- 6 years, 3 months ago

Wolfram | Alpha

- 6 years, 3 months ago

Why not use integration by parts

- 6 years, 3 months ago

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

- 6 years, 3 months ago