Mathematics

# $\displaystyle \underset{1}{\overset{2}{\int}} \dfrac{1 - x}{1 + x} dx$ equals

$2 \, \log \, \left(\dfrac{3}{2} \right) - 1$

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

$u=1+x\rightarrow du=dx$

$=\int \dfrac{-u+2}{u}du$

$=\int -1+\dfrac{2}{u}du$

$=-u+2\log |u| +C$

$=-(1+x)+2\log |(1+x)| +C$

$\int_{1}^{2}\dfrac{1-x}{1+x}dx=(2\log 3 -3) -(2\log 2-2)$

$=\log \left ( \dfrac{9}{4} \right )-1$

$=\log \left ( \dfrac{3}{2} \right )^2-1$

$=\left ( 2\right ) \log \left ( \dfrac{3}{2} \right )-1$

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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
$\displaystyle \int\frac{a}{b+ce^{x}}dx$ is equal to
• A. $\displaystyle \frac{a}{b}\log|b+c\cdot e^{x}|+K$
• B. $\log|b+c\cdot e^{x}|+K$
• C. $\log|c+b\cdot e^{-x}|+K$
• D. $-\displaystyle \frac{a}{b}\log|c+be^{-x}|+K$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate the given integral.
$\displaystyle \int { \cfrac { { e }^{ x }\left( 1+x \right) }{ \cos ^{ 2 }{ \left( x{ e }^{ x } \right) } } } dx$
• A. $2\log _{ e }{ \cos { \left( x{ e }^{ x } \right) } } +C$
• B. $\sec { \left( x{ e }^{ x } \right) +C }$
• C. $\tan { \left( x+{ e }^{ x } \right) } +C$
• D. $\tan { \left( x{ e }^{ x } \right) } +C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Solve :
$\displaystyle I = \int \dfrac { \cos x}{ 2 \sin^2 x + 3 \sin x + 1 } dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Multiple Correct Medium
If $I=\sum _{ k=1 }^{ 98 }{ \displaystyle\int _{ k }^{ k+1 }{ \cfrac { k+1 }{ x(x+1) } } } dx$, then
• A. $I< \log _{ e }{ 99 }$
• B. $I< \cfrac{49}{50}$
• C. $I> \log _{ e }{ 99 }$
• D. $I> \cfrac{49}{50}$

$\displaystyle \overset{2}{\underset{3}{\int}} \dfrac{x^3 + 1}{x(x - 1)}dx$.