Mathematics

# $\displaystyle \int \frac{e^{x}-1}{e^{x}+1}dx.$

$\displaystyle 2\log \left ( e^{x}+1 \right )-\log e^{x}$

##### SOLUTION
Let $\displaystyle I=\int \frac { e^{ x }-1 }{ e^{ x }+1 } dx=\int \frac { e^{ x }\left( e^{ x }-1 \right) }{ e^{ x }\left( e^{ x }+1 \right) } dx$

Put $\displaystyle e^{ x }=t\Rightarrow e^{ x }dx=dt$

$\displaystyle I=\int \frac { t-1 }{ t\left( t+1 \right) } dt=\int \left( \frac { 2 }{ t+1 } -\frac { 1 }{ t } \right) dt$

$\displaystyle =2\log \left( t+1 \right) -\log t=2\log \left( e^{ x }+1 \right) -\log e^{ x }$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
Evaluate: $\int \displaystyle \frac{\sin^{2} x}{\cos^{4} x}dx$
• A. $\displaystyle \frac{1}{2} \tan^{2}\, x + c$
• B. $\displaystyle \frac{1}{2} \cot^{2}\, x + c$
• C. $\displaystyle \frac{1}{3} \cot^{3}\, x + c$
• D. $\displaystyle \frac{1}{3} \tan^{3}\, x + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\displaystyle \int \left( \frac { x^6 -1 }{x^2 +1 } \right) dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Resolve into partial fractions $\displaystyle \frac{x^3}{(x-1)^4(x^2-x+1)}$
• A. $\displaystyle \frac{1}{(x-1)^4}+\frac{2}{(x-1)^2}+\frac{x}{x^2-x+1}$
• B. $\displaystyle \frac{1}{(x-1)^4}-\frac{1}{(x-1)^2}-\frac{1}{(x-1)}+\frac{x}{x^2-x+1}$
• C. $\displaystyle \frac{-1}{(x-1)^4}+\frac{2}{(x-1)^2}-\frac{1}{(x-1)}+\frac{x}{x^2+x+1}$
• D. $\displaystyle \frac{1}{(x-1)^4}+\frac{2}{(x-1)^2}-\frac{1}{(x-1)}+\frac{x}{x^2-x+1}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Evaluate: $\underset {n \rightarrow \infty} \lim \displaystyle \sum_{r=0}^{n-1}\frac{1}{n+r}$
• A. $2 \log 2$
• B. $\dfrac{1}{2} \log 2$
• C. $\dfrac{1}{4} \log 2$
• D. $\log 2$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$