Mathematics

# $\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } }$

##### SOLUTION
$\displaystyle \int _{-1}^{1}\dfrac {dx}{x^2+zx+5}$
$\displaystyle \int _{-1}^{1}\dfrac {1}{(x+1)^2+4}dx$
$\dfrac {1}{2}\tan^{-1}\dfrac {(x+1)}{2}|_{-1}^{1}$
$\dfrac {1}{2}\tan^{-1} \left (\dfrac {2}{2}\right)-\dfrac {1}{2}\tan^{-1}\left (\dfrac {-1+1}{2}\right)$
$\dfrac {1}{2}\tan^{-1} 1-\dfrac {1}{2}\tan^{-1}0$
$\dfrac {1}{2} \left (\dfrac {\pi}{4}\right)-\dfrac {1}{2}(0)\ \Rightarrow \dfrac {\pi}{8}-0=\boxed {\dfrac {\pi}{8}}$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Single Correct Hard
$\int \dfrac{cos 2x - cos 2 \theta}{cos x - cos \theta} dx$ is equal to
• A. $2(sin x - x cos \theta) + C$
• B. $2(sin x + 2x cos \theta) + C$
• C. $2(sin x - 2x cos \theta) + C$
• D. $2(sin x + x cos \theta) + C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Let $f(x)$ be a function satisfies ${f}'\left ( x \right )=f\left ( x \right )$ with $f(0)=1$ and $g(x)$ be a function that satisfies $f\left ( x \right )+g\left ( x \right )=x^{2}$, then the value of the integral $\int_{0}^{1}f\left ( x \right )g\left ( x \right )dx$ is
• A. $\displaystyle e+\frac{e^{2}}{2}-\frac{3}{2}$
• B. $\displaystyle e+\frac{e^{2}}{2}+\frac{5}{2}$
• C. $\displaystyle e-\frac{e^{2}}{2}-\frac{5}{2}$
• D. $\displaystyle e-\frac{e^{2}}{2}-\frac{3}{2}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate the following integral:
$\displaystyle \int { \cfrac { \left( x+1 \right) { e }^{ x } }{ \sin ^{ 2 }{ \left( x{ e }^{ x } \right) } } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle\int \frac{2x+\sin 2x}{1+\cos 2x}dx$
• A. $x\cot x.$
• B. $x^{2}\tan x.$
• C. $x\sec x.$
• D. $x\tan x.$

Evaluate $\displaystyle \int_{0}^{\pi/2} cos \,x \,e^{sin \,x} \,dx$.