Mathematics

# If $\displaystyle \int \dfrac{e^x - 1}{e^x + 1}dx =f(x) + c$ then $f(x) =$

##### SOLUTION
$\displaystyle \int \dfrac{e^x - 1}{e^x + 1}dx$

$=\displaystyle \int \dfrac{e^x}{e^x + 1}dx $$-\displaystyle \int \dfrac{ 1}{e^x + 1}dx =\displaystyle \int \dfrac{e^x}{e^x + 1}dx$$-\displaystyle \int \dfrac{ e^{-x}}{e^{-x} + 1}dx$ [Multiplying the numerator and denominator of the second integral by $e^{-x}$]

$=\displaystyle \int \dfrac{d(e^x+1)}{e^x + 1}dx$$+\displaystyle \int \dfrac{d( e^{-x}+1)}{e^{-x} + 1}dx$
$=\log(1+e^x)+\log(1+e^{-x})+c$ [$c$ being integrating constant]
$=\log(1+e^x)+\log(1+e^{x})-x+c$
$=2\log(1+e^x)-x+c$
Comparing this with the given integral we get,
$f(x)=2\log(1+e^{x})-x$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Single Correct Hard
$\displaystyle\int{\frac{(1+x^2)dx}{(1-x^2)\sqrt{1+x^2+x^4}}} =$
• A. $\displaystyle I=-\frac{1}{2\sqrt{3}}\log{\left|\frac{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}-\sqrt{5}}{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}+\sqrt{5}}\right|}+C$
• B. $\displaystyle I=-\frac{1}{4\sqrt{3}}\log{\left|\frac{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}-\sqrt{3}}{\sqrt{2x^2+\displaystyle\frac{1}{x^2}+1}+\sqrt{3}}\right|}+C$
• C. $\displaystyle I=-\frac{1}{2\sqrt{3}}\log{\left|\frac{\sqrt{2x^2+\displaystyle\frac{1}{2x^2}+1}-\sqrt{3}}{\sqrt{2x^2+\displaystyle\frac{1}{2x^2}+1}+\sqrt{3}}\right|}+C$
• D. $\displaystyle I=-\frac{1}{2\sqrt{3}}\log{\left|\frac{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}-\sqrt{3}}{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}+\sqrt{3}}\right|}+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\int_{0}^{2}\dfrac{dx}{(17+8x-4x^2)(e^{6(1-x)}+1)}$ is equal to
• A. $-\dfrac{1}{8\sqrt{21}}\log \left | \dfrac{2-\sqrt{21}}{2+\sqrt{21}} \right |$
• B. $-\dfrac{1}{8\sqrt{21}}\log \left | \dfrac{2+\sqrt{21}}{\sqrt{21}-2} \right |$
• C. $-\dfrac{1}{8\sqrt{21}}\left \{ \log \left | \dfrac{2-\sqrt{21}}{2+\sqrt{21}}\right |-\log \left | \dfrac{2+\sqrt{21}}{\sqrt{21}-2} \right | \right \}$
• D. None of these

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int { \frac { \cos { x } -\sin { x } }{ \sqrt { 8-\sin { 2x } } } dx }$ is equal to
• A. $\sin ^{ -1 }{ \left( \sin { x } +\cos { x } \right) } +c$
• B. $\cos ^{ -1 }{ \left( \sin { x } +\cos { x } \right) } +c$
• C. None of these
• D. $\displaystyle \sin ^{ -1 }{ \left[ \frac { 1 }{ 3 } \left( \sin { x } +\cos { x } \right) \right] } +c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Evaluate $\displaystyle \int\frac{1}{\sin^{3}x\cos^{5}x}dx$
• A. $3$ $\displaystyle \log\tan x+\frac{3}{2}\tan^{2}x+\frac{1}{4}\tan^{4}x-\frac{1}{2}\tan^{2}x+c$
• B. $3\displaystyle \log\cot x+\frac{3}{2}\cot^{2}x+\frac{1}{4}\cot^{4}x-\frac{1}{2}\tan^{2}x+c$
• C. $3\displaystyle \log\cot x+\frac{3}{2}\cot^{2}x+\frac{1}{4}\cot^{4}x-\frac{1}{2}\cot^{2}x+c$
• D. $3\displaystyle \log tanx+\frac{3}{2}\tan^{2}x+\frac{1}{4}\tan^{4}x-\frac{1}{2}\cot^{2}x+c$

$\int \frac{2x^{2}}{3x^{4}2x} dx$