Mathematics

# If $\int _0^1 xdx = \dfrac {\pi}{4} - \dfrac {1}{2} ln 2$ then the value of definite integral $\int _0^1 \tan^{-1} (1-x+x^2) dx$ equals :

$ln2$

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

#### Realted Questions

Q1 Single Correct Medium
Evaluate the integral
$\displaystyle \int_{0}^{1}\frac{1}{1+x^{2}}dx$
• A. $\pi$
• B. $\pi/3$
• C. $0$
• D. $\pi/4$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Integrate  with respect to x : $\dfrac{\cos 2x - \cos 2 \alpha}{\cos x - \cos \alpha}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\int_{-3\pi /2}^{-\pi /2}\left [ \left ( x+\pi \right )^{3}+\cos ^{2}\left ( x+3\pi \right ) \right ]dx$ is equal to
• A. $\displaystyle \frac{\pi ^{4}}{32}$
• B. $\displaystyle \frac{\pi ^{4}}{32}+\frac{\pi }{2}$
• C. $\displaystyle \frac{\pi }{2}-1$
• D. $\displaystyle \frac{\pi }{2}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int_{1}^{e}\log x \,dx =$________
• A. $e + 1$
• B. $e -1$
• C. $e + 2$
• D. $1$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } \right) } dx }$
• A. $\dfrac{\pi}{2}$
• B. $0$
• C. None of these
• D. $\dfrac{\pi}{2}-1$