Mathematics

# Evaluate the following definite integral:$\displaystyle \int_{-1}^{1}\dfrac {1}{x^2 +2x+5}dx$

##### SOLUTION
Consider, $I=\displaystyle \int_{-1}^1\dfrac {1}{x^2+2x+5}dx$

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

$I=\dfrac {1}{2}\left [\tan^{-1} \dfrac {x+1}{2}\right]_{1-}^1$

$I=\dfrac {1}{2}(\tan^{-1} 1-\tan^{-1}0)$

$I=\dfrac {\pi}{8}$

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int { \frac { dx }{ \left( 1+\sqrt { x } \right) \sqrt { x-{ x }^{ 2 } } } }$ is equal to
• A. $\displaystyle \frac { 2\left( -\sqrt { x } +1 \right) }{ \sqrt { x-1 } } +c$
• B. $\displaystyle \frac { \left( \sqrt { x } -1 \right) }{ \sqrt { x-1 } } +c$
• C. None of these
• D. $\displaystyle \frac { 2\left( \sqrt { x } -1 \right) }{ \sqrt { x-1 } } +c$

1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Hard
$\displaystyle \int \frac{\sqrt{1+x^{2n}}\left \{ \log \left ( 1+x^{2n} \right ) -2n\log x\right \}}{x^{3n+1}}dx=-\frac{1}{2n}\int \sqrt{t}.\log t dt=-\frac{1}{2n}\left [ \frac{2}{3}t\sqrt{t}\log t-\frac{4}{9}t\sqrt{t} \right ]$ where $t=1+\frac{1}{x^{2n}}$. If this is true enter 1, else enter 0.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of
$\displaystyle \int{(1+(2tanx)(\tan x+\sec x))}^{1/2}$ dx  is equal to
• A. $In\|sec x(\csc x\tan x)|+C$
• B. $In\|sec x(\sec x\tan x)|+C$
• C. $In\|cos x(\sec x\tan x)|+C$
• D. $In\|sec x(\sec x+\tan x)|+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Evaluate: $\displaystyle \int_{0}^{\pi/2}\log (\tan x) \ dx$
• A. $\displaystyle \frac{\pi}{4}$
• B. $\displaystyle \frac{\pi}{2}$
• C. $\pi$
• D. $0$

Solve: $\int {{{\left( {x + \dfrac{1}{x}} \right)}^3}dx}$