Question

integrate

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

Answer 1

To Integrate :

$\star \int \frac{ {e}^{ { \tan}^{ - 1} x} }{1 + {x}^{2} } dx$

Here ,

• we use the substitution method :

Let ,

$\star \: { \tan}^{ - 1} x = t$

Differentiate w.r.t x both sides ,

$\star \: \frac{1}{1 + {x}^{2} } = \frac{dt}{dx} \\ \\ \star \: dx = (1 + {x}^{2} )dt$

put the values :

$\implies \: \int \frac{ {e}^{t} }{1 + {x}^{2} } (1 + {x}^{2} )dt \\ \\ \implies \: \int \: {e}^{t} dt \\ \\ \implies \: {e}^{t} + c$

Now, put the value of t .

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

Formula :

$\star \bold{ \frac{d}{dx} {tan}^{ - 1} x \: = \frac{1}{1 + {x}^{2} } } \\ \\ \star \bold{\int \: {e}^{x} dx = {e}^{x} + c}$

Answer 2

To Integrate :

\begin{lgathered}\star \int \frac{ {e}^{ { \tan}^{ - 1} x} }{1 + {x}^{2} } dx \\\end{lgathered}

⋆∫

1+x

2

e

tan

−1

x

dx

Here ,

• we use the substitution method :

Let ,

\star \: { \tan}^{ - 1} x = t⋆tan

−1

x=t

Differentiate w.r.t x both sides ,

\begin{lgathered}\star \: \frac{1}{1 + {x}^{2} } = \frac{dt}{dx} \\ \\ \star \: dx = (1 + {x}^{2} )dt\end{lgathered}

⋆

1+x

2

1

=

dx

dt

⋆dx=(1+x

2

)dt

put the values :

\begin{lgathered}\implies \: \int \frac{ {e}^{t} }{1 + {x}^{2} } (1 + {x}^{2} )dt \\ \\ \implies \: \int \: {e}^{t} dt \\ \\ \implies \: {e}^{t} + c\end{lgathered}

⟹∫

1+x

2

e

t

(1+x

2

)dt

⟹∫e

t

dt

⟹e

t

+c

Now, put the value of t .

\begin{lgathered}\implies \: {e}^{ { \tan}^{ - 1}x } + c \\\end{lgathered}

⟹e

tan

−1

x

+c

Formula :

\begin{lgathered}\star \bold{ \frac{d}{dx} {tan}^{ - 1} x \: = \frac{1}{1 + {x}^{2} } } \\ \\ \star \bold{\int \: {e}^{x} dx = {e}^{x} + c}\end{lgathered}

⋆

dx

d

tan

−1

x=

1+x

2

1

⋆∫e

x

dx=e

x

+c