Mathematics

# $\int _{ -1 }^{ 1 }{ x } \tan { ^{ -1 } } xdx$

$\left( \frac { \pi }{ 2 } -1 \right)$

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

#### Realted Questions

Q1 Subjective Medium
Find: $\displaystyle \int {e^{y}(y^2+2y)} d y$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\int { { x }^{ 3 }\log { x } dx }$ is
• A. $\dfrac { 1 }{ 8 } \left( { x }^{ 4 }\log { x } -4{ x }^{ 4 }+c \right)$
• B. $\dfrac { 1 }{ 16 } \left( 4{ x }^{ 4 }\log { x } +{ x }^{ 4 }+c \right)$
• C. $\dfrac { { x }^{ 4 }\log { x } }{ 4 } +c$
• D. $\dfrac { 1 }{ 16 } \left( 4{ x }^{ 4 }\log { x } -{ x }^{ 4 }+c \right)$

1 Verified Answer | Published on 17th 09, 2020

Q3 TRUE/FALSE Medium
We can integrate$\dfrac{3x+5}{x^2+4x+13}$using partial fractions.
• A. True
• B. False

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int\frac{dx}{1+e^{-x}}=$
• A. $1+e^{x}+c$
• B. $\dfrac{1}{2}+(1+e^{x})+c$
• C. $\dfrac{1}{2}\log(1+e^{x})+c$
• D. $\log(1+e^{x})+c$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\displaystyle \int\frac{dt}{(6t-1)}$ is equal to:
• A. $\ln(6t-1)+C$
• B. $-\dfrac{1}{6}\ln(6t-1)+C$
• C. None of these
• D. $\dfrac{1}{6} \ln(6t-1) +C$