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..

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

Why not use integration by parts

Log in to reply

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

Log in to reply

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

Log in to reply

HOW???

Log in to reply

Wolfram | Alpha

Log in to reply

You should check this.

Log in to reply

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

Log in to reply

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?)

Log in to reply

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..

Log in to reply

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

Log in to reply