Question

integral of e^(1-1/x)* (1+1/x^2)​

Answer 1

Appropriate Question :- Evaluate

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

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

Given integral is

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

To solve this integral, we use method of Substitution.

So, substitute

$\rm \: x - \dfrac{1}{x} = y$

can be rewritten as

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

On differentiating both sides w. r. t. x, we get

$\rm \: 1 + {x}^{ - 1 - 1} = \dfrac{dy}{dx}$

$\rm \: \bigg(1 + \dfrac{1}{ {x}^{2} } \bigg)dx \: = \: dy$

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

$\rm \: = \: \displaystyle\int\rm {e}^{y} \: dy$

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

$\rm \: = \: {e}^{ \: \bigg(x - \dfrac{1}{x} \bigg) } \: + \: c$

Hence,

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

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

Additional Information :-

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