Question

Please answer .
Jawab ata ho to do nahi to apni ungliyo ko kasht mat dena
​

Answer 1

AnswEr :

Given Expression,

$\displaystyle \sf \: l = \int \dfrac{dx}{ {e}^{x} - 1}$

Multiplying and Dividing by e^x + 1, we get :

$\implies \displaystyle \sf \: l = \int \dfrac{dx}{ {e}^{x} - 1} \times \dfrac{ {e}^{x} }{e {}^{x} } \\ \\ \implies \displaystyle \sf \: l = \int \dfrac{ {e}^{x}}{ {e}^{2x} - {e}^{x} } dx$

Let e^x = u.

Differentiating w.r.t x on both sides,

$\longrightarrow \sf \: \dfrac{du}{dx} = \dfrac{d( {e}^{x} )}{dx} \\ \\ \longrightarrow \sf \: dx = \dfrac{du}{e {}^{x} }$

Therefore,

$\implies \displaystyle \: \sf \: l = \int \dfrac{ {e}^{x}}{ {u}^{2} - u } \times \dfrac{du}{e {}^{x} } \\ \\ \implies \displaystyle \: \sf \: l = \int \dfrac{u - (u - 1)}{ {u}^{2} - u } du \\ \\ \implies \displaystyle \: \sf \: l = \int \dfrac{du}{u - 1} - \int \dfrac{du}{u} \\ \\ \implies \sf \: l = log(u - 1) - log_{}(u)$

We know that,

log(a) - log(b) = log(a/b)

Thus,

$\implies \sf l = log \bigg( \dfrac{u - 1}{u} \bigg) \\ \\ \sf \implies l = log \bigg( \dfrac{e {}^{x} - 1}{ {e}^{x} } \bigg) \\ \\ \implies \boxed{ \boxed{\sf l = log |1 - {e}^{ - x} | + c}}$

Option (3) is correct.

Answer 2

Step-by-step explanation:

AnswEr :

Given Expression,

\displaystyle \sf \: l = \int \dfrac{dx}{ {e}^{x} - 1}l=∫

e

x

−1

dx

Multiplying and Dividing by e^x + 1, we get :

\begin{gathered} \implies \displaystyle \sf \: l = \int \dfrac{dx}{ {e}^{x} - 1} \times \dfrac{ {e}^{x} }{e {}^{x} } \\ \\ \implies \displaystyle \sf \: l = \int \dfrac{ {e}^{x}}{ {e}^{2x} - {e}^{x} } dx\end{gathered}

⟹l=∫

e

x

−1

dx

×

e

x

e

x

⟹l=∫

e

2x

−e

x

e

x

dx

Let e^x = u.

Differentiating w.r.t x on both sides,

\begin{gathered} \longrightarrow \sf \: \dfrac{du}{dx} = \dfrac{d( {e}^{x} )}{dx} \\ \\ \longrightarrow \sf \: dx = \dfrac{du}{e {}^{x} } \end{gathered}

⟶

dx

du

=

dx

d(e

x

)

⟶dx=

e

x

du

Therefore,

\begin{gathered}\implies \displaystyle \: \sf \: l = \int \dfrac{ {e}^{x}}{ {u}^{2} - u } \times \dfrac{du}{e {}^{x} } \\ \\ \implies \displaystyle \: \sf \: l = \int \dfrac{u - (u - 1)}{ {u}^{2} - u } du \\ \\ \implies \displaystyle \: \sf \: l = \int \dfrac{du}{u - 1} - \int \dfrac{du}{u} \\ \\ \implies \sf \: l = log(u - 1) - log_{}(u) \end{gathered}

⟹l=∫

u

2

−u

e

x

×

e

x

du

⟹l=∫

u

2

−u

u−(u−1)

du

⟹l=∫

u−1

du

−∫

u

du

⟹l=log(u−1)−log

(u)

We know that,

log(a) - log(b) = log(a/b)

Thus,

\begin{gathered}\implies \sf l = log \bigg( \dfrac{u - 1}{u} \bigg) \\ \\ \sf \implies l = log \bigg( \dfrac{e {}^{x} - 1}{ {e}^{x} } \bigg) \\ \\ \implies \boxed{ \boxed{\sf l = log |1 - {e}^{ - x} | + c}}\end{gathered}

⟹l=log(

u

u−1

)

⟹l=log(

e

x

e

x

−1

)

⟹

l=log∣1−e

−x

∣+c

Option (3) is correct.