Question

Integrate the given function with respect to x

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

Answer 1

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

Given integral is

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

To solve 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

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

We know,

$\boxed{\tt{ {sec}^{2}x - {tan}^{2}x = 1 \: }}$

So, using this identity, we get

$\rm \: = \: \displaystyle\int\rm {e}^{y}\bigg[\dfrac{ tany + {sec}^{2}y }{ {sec}^{2}y} \bigg] \: {sec}^{2}y \: dy$

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

We know,

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

So, here

$\rm \: f(x) = tanx$

and

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

So, using above result, we get

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

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

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

Hence,

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

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ADDITIONAL INFORMATION

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

Answer 2

Question:-

Integrate the given function with respect to x

$\sf \large {e}^{ {tan}^{ - 1} x} ( \frac{1 + x + {x}^{2} }{1 + {x}^{2} } ).$

Given:-

$\sf \large \int {e}^{ {tan}^{ - 1} x} ( \frac{1 + x + {x}^{2} }{1 + {x}^{2} } )$

To Find:-

• The value of x.

Solution:-

$\sf \large Let, I = \int e {}^{tan - 1 \: x} \bigg [ \frac{1 + x + x {}^{2} }{1 + x {}^{2} } \bigg ] dx$

$\sf \large = \int e {}^{tan - 1 \: x} \bigg [ \frac{1 + x {}^{2} }{1 + x {}^{2} } + \frac{x}{1 + x {}^{2} } \bigg ]dx$

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

$\sf \large = \int e {}^{tan - 1 \: x} .1 \: dx + \int e {}^{tan - 1 \: x} ( \frac{x}{1 + x {}^{2} } )dx$

$\sf \large = e {}^{tan - 1 \: x } \int1 \: dx - \int \bigg[ \frac{d}{dx} \: e {}^{tan - 1 \: x} \int 1 \: dx \bigg ] + \int e {}^{tan - 1 \: x} ( \frac{x}{1 +x {}^{2} } )dx$

$\sf \large = x.e {}^{tan - 1 \: x} - \int \frac{e {}^{tan - 1 \: x} }{1 + x {}^{2} } \: x \: dx + \int e {}^{tan - 1 \: x}$

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

Answer:-

${ \boxed{ \sf \large \green{\sf \large \int {e}^{ {tan}^{ - 1} x} ( \frac{1 + x + {x}^{2} }{1 + {x}^{2} } )= x \: e {}^{tan - 1 \: x} + c.}}}$

Hope you have satisfied. ⚘