Mathematics

# If $2\displaystyle \int_{0}^{1}\tan^{-1}xdx=\displaystyle \int_{0}^{1}\cot^{-1}(1-x+x^{2})dx$, then $\displaystyle \int_{0}^{1}\tan^{-1}(1-x+x^{2})dx$ is equal to:

$\log 4$

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

#### Realted Questions

Q1 Subjective Medium
Find:
$\displaystyle \int \dfrac {xe^{x}}{(1+x)^{2}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$I = \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{\sin x}}{x}dx}$, then
• A. $\frac{1}{2} \le I \le 1$
• B. $4 \le I \le 2\sqrt {30}$
• C. $1 \le I \le \frac{{2\sqrt 3 }}{{\sqrt 2 }}$
• D. $\frac{{\sqrt 3 }}{8} \le I \le \frac{{\sqrt 2 }}{6}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int\frac{e^{x}}{(2e^{x}-3)^{2}}dx=$
• A. $\displaystyle \frac{1}{2e^{x}-3}+c$
• B. $\displaystyle \frac{1}{2(2e^{x}-3)^{2}}+c$
• C. $\displaystyle \frac{1}{(2e^{x}-3)^{2}}+c$
• D. $\displaystyle \frac{-1}{2(2e^{x}-3)}+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Solve $\int x ^ { 2 } \cos x dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\int { \dfrac { dx }{ \sqrt { 9x+4{ x }^{ 2 } } } }$ equals
• A. $\dfrac { 1 }{ 2 } \sin ^{ -1 }{ \left( \dfrac { 8x-9 }{ 9 } \right) +C }$
• B. $\dfrac { 1 }{ 3 } \sin ^{ -1 }{ \left( \dfrac { 9x-8 }{ 8 } \right) +C }$
• C. $\dfrac { 1 }{ 2 } \sin ^{ -1 }{ \left( \dfrac { 9x-8 }{ 9 } \right) +C }$
• D. $\dfrac { 1 }{ 9 } \sin ^{ -1 }{ \left( \dfrac { 9x-8 }{ 8 } \right) +C }$