Mathematics

# Evaluate$\int\limits_0^\infty {\frac{{\log \left( {1 + {x^2}} \right)dx}}{{1 + {x^2}}}} =$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
Solve $\displaystyle\int { \frac { 1 }{ \sqrt { 3x+4 } -\sqrt { 3x+1 } } dx }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int \:\dfrac{1-x^7}{x\left(1+x^7\right)}dx$ equals
• A. $ln\left|x\right|-\dfrac{2}{4}ln\left|1-x^7\right|+c$
• B. $ln\left|x\right|-\dfrac{2}{7}ln\left|1+x^7\right|+c$
• C. $ln\left|x\right|+\dfrac{2}{4}ln\left|1-x^7\right|+c$
• D. $ln\left|x\right|+\dfrac{2}{7}ln\left|1+x^7\right|+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate the integral
$\displaystyle \int_{0}^{1}\tan^{-1}x \ dx$
• A. $\dfrac{\pi}{4}-\dfrac{1}{4} log2$
• B. $\displaystyle \frac{\pi}{4}+\frac{1}{2} log 2$
• C. $\displaystyle \frac{\pi}{4}+\frac{1}{4} log 2$
• D. $\displaystyle \frac{\pi}{4}-\frac{1}{2} log2$

1 Verified Answer | Published on 17th 09, 2020

Q4 One Word Hard
The value of the integral
$\displaystyle \int_{0}^{\frac {1}{2}}\frac {1 + \sqrt {3}}{(x + 1)^{2} (1 - x)^{6})^{\frac {1}{4}}}dx$ is _______.

If $\displaystyle \int \left[\left(\dfrac{x}{e}\right)^x + \left(\dfrac{e}{x} \right)^x \right] \ln x \, dx = A \left(\dfrac{x}{e}\right)^x + B \left(\dfrac{e}{x} \right)^x + C$, then value of $A + B =$?