Question

integrate e- tan-1x (1+x+x 2 /1+x 2 )dx

Answer 1

Hope it is clear to you.

Answer 2

Answer:

$= x \times e^{tan^{-1} x}$

Step-by-step explanation:

We have been given to integrate the following :

$e^{tan^{-1}x } \times (\frac{1 +x + x^{2} }{1+x^{2} } )$$\int\limits^ {} [tex]e^{tan^{-1}tant } \,  \times \frac{(1 + tant + tan^{2}t) }{1 + tan^{2}t } dt$

Let us assume that :

$x = tan t$

$dx = sec^{2} t dt$

$e^{tan^{-1}tant } \,  \times \frac{(1 + tant + tan^{2}t) }{1 + tan^{2}t } dt$

$\int\limits^ {} \, e^{t} \times\frac{1 + tant + tam^{2}t }{1 + tan^{2}t } \times sec^{2} t dt$

$e^{t}  \times (\frac{sec^{2}t + tant  }{sec^{2}t } ) \times sec^{2} t dt$

$e^{t} sec^{2} t + e^{t} tant$

Now, evaluating by parts:

$e^{t} tant = tant \times\int\limits^{} e^{t} \, dt - \int\limits^ {}(sec^{2} t \, \int\limits^{} e^{t} dt)\, dt</p><p>[tex]e^{t} tant = tant \times\int\limits^{} e^{t} \, dt - \int\limits^ {}(sec^{2} t \, \int\limits^{} e^{t} dt)\, dt + e^{t} sec^{2} t dt$

$= e^{t} tant$

Now , put t = $tan^{-1} x$

∴ $e^{tan^{-1}x } \times tantan^{-1} x$

$= x \times e^{tan^{-1} x}$

∴ The final answer will be $= x \times e^{tan^{-1} x}$