Question

Solve the integration

$\int \: {e}^{ {tan}^{ - 1} x}( \frac{1 + x + {x}^{2} }{1 + {x}^{2} }) \: dx$
​

Answer 1

Answer:

I = e^(tan-¹x) x + c

Step-by-step explanation:

I = ∫ e^(tan-¹x) (1 + x + x²/ 1 + x²) dx

Take tan-¹ x = t ----> (1)

Now differentiate the above equation to obtain the following :

(1/1 + x²)dx = dt -----> (2)

Now we can substitute (2) in the original question by re arranging the terms a bit and hence having a simpler integrand.

Using (1),

I = ∫ e^(t) (1 + x + x²/1 + x²) dx

Now,for some ease,

I = ∫ e^(t) (1 + x + x² × [1/1 + x²]) dx

See the term in the square bracket, it's the same as (2),so we can write dt there instead of that complicated term,

I = ∫ e^(t) (1 + x + x²) dt

Also, using (1), tan-¹x = t => x = tant,

I = ∫ e^(t) (1 + tant + tan²t) dt

I = ∫ e^(t) ( tant + 1 + tan²t) dt

Globally it is known that, 1 + tan²t = sec²t,

I = ∫ e^(t) (tant + sec²t) dt ---> (A)

Now, if you have practice for some nice hours, you will surely identify what's happening here. The property if you have some function inside the int with e^(x)•f(x) and also the derivative of f(x) i.e f'(x) is present beside it (something like this ∫ e^(x) [ f(x) + f'(x)])) then the direct answer is : e^(x) f(x) + c.

So,if you observe (A), you can clearly see that the derivative of tant is present with it i.e sec²t,so :

I = e^(t) tant + c

Now just resubstitute all the initial variables and powers,

I = e^(tan-¹x) x + c

You can solve it with a different approach as well,but it will just consume your time,so you must be knowing all these properties, they can be really handy at times.

Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\rm {e}^{ {tan}^{ - 1} x}\bigg(\dfrac{1 + x + {x}^{2} }{1 + {x}^{2} } \bigg) \: dx$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\rm \: {tan}^{ - 1}x = y$

$\rm \: x = tany$

$\rm \: dx = {sec}^{2}y \: dy$

So, on substituting these values in above integral, we get

$\rm \: = \: \displaystyle\int\rm {e}^{y}\bigg(\dfrac{1 + tany + {tan}^{2}y}{1 + {tan}^{2} y} \bigg) \: {sec}^{2}y \: dy$

$\rm \: = \: \displaystyle\int\rm {e}^{y}\bigg(\dfrac{1+ {tan}^{2}y + tany}{{sec}^{2} y} \bigg) \: {sec}^{2}y \: dy$

$\rm \: = \: \displaystyle\int\rm {e}^{y}( {sec}^{2}y + tany) \: dy$

We know,

$\boxed{ \rm{ \: \displaystyle\int\rm {e}^{x}[f(x) + f'(x)] \: dx\: = \: {e}^{x} \: f(x) \: + \: c \: }}$

So, here

$\rm \: f(y) = tany$

$\rm \: f'(y) = {sec}^{2}y$

So, using this, we get

$\rm \: = \: {e}^{y} \: tany \: + \: c$

$\rm \: = \: {e}^{ {tan}^{ - 1} x} \: x \: + \: c$

So,

$\boxed{ \rm{ \:\rm \: \displaystyle\int\rm {e}^{ {tan}^{ - 1} x}\bigg(\dfrac{1 + x + {x}^{2} }{1 + {x}^{2} } \bigg) \: dx = x \: {e}^{ {tan}^{ - 1} x} + c \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

Proof of result :-

$\boxed{ \rm{ \: \displaystyle\int\rm {e}^{x}[f(x) + f'(x)] \: dx\: = \: {e}^{x} \: f(x) \: + \: c \: }}$

Consider,

$\displaystyle\int\rm {e}^{x}[f(x) + f'(x)] \: dx$

$\rm \: = \: \displaystyle\int\rm {e}^{x} \: f(x) \: dx +\displaystyle\int\rm {e}^{x} f'(x) \: dx$

By using by parts in first integral, we get

$\rm \: = \:f(x) \displaystyle\int\rm {e}^{x} \: dx - \displaystyle\int\rm \bigg[\dfrac{d}{dx}f(x)\displaystyle\int\rm {e}^{x} dx\bigg] dx+\displaystyle\int\rm {e}^{x} f'(x) \: dx$

$\rm \: = \: {e}^{x} f(x) - \displaystyle\int\rm f'(x){e}^{x} dx \: + c + \displaystyle\int\rm {e}^{x} f'(x) \: dx$

$\rm \: = \: {e}^{x} f(x) \: + \: c$

$\rule{190pt}{2pt}$

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$