Mathematics

# Evaluate: $\displaystyle \int_{0}^{2}\dfrac {1}{4+x-x^2}dx$

##### SOLUTION
$I=\displaystyle \int_0^2 \dfrac {1}{4+x-x^2}dx$

$I=-\displaystyle \int_0^2 \dfrac {1}{x^2 -x-4}dx$

$I=-\displaystyle \int_0^2\dfrac {1}{\left (x-\dfrac {1}{2}\right)^2-\left (\dfrac {\sqrt {17}}{2}\right)^2}dx$

$\ I=\displaystyle \int_0^2 \dfrac {1}{\left (\dfrac {\sqrt {17}}{2}\right)^2-\left (x-\dfrac {1}{2}\right)^2}dx$

$I=\dfrac {1}{\sqrt {17}} \left [\log \left (\dfrac {\sqrt {17}+2x-1}{\sqrt {17} -2x+1}\right) \right]_0^2$

$\ I=\dfrac {1}{\sqrt {17}}\left\{\log \dfrac {\sqrt {17}+3}{\sqrt {17}-3} -\log \dfrac {\sqrt {17}-1}{\sqrt {17}+1} \right\}$

$I=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {20+4\sqrt {17}}{20-4\sqrt {17}} \times \dfrac {20+4\sqrt {17}}{20+4\sqrt {17}}\right)$

$I=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {672+160\sqrt {17}}{128}\right)$

$I=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {21+5\sqrt {17}}{4}\right)$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Subjective Medium
Evaluate the following integral:
$\displaystyle \int { co\sec { x } \log { \left( co\sec { x } -\cot { x } \right) } } dx\quad$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\int sin^{2/3}x cos^{3}x dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
$\int _{ 0 }^{ \pi /2 }{ x\sin { x } \cos { x } dx }$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Solve:
$\displaystyle \int_{0}^{2\pi}{e^{x}.\sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)dx}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\int { (\cfrac { { 2 }^{ x }-5^{ x } }{ 10^{ x } } )dx }$ is equal to __________________.
• A. $\cfrac { { 2 }^{ x } }{ \log _{ e }{ 2 } } -\cfrac { 5^{ x } }{ \log _{ e }{ 5 } } +c$
• B. $\cfrac { { 2 }^{ x } }{ \log _{ e }{ 2 } } +\cfrac { 5^{ x } }{ \log _{ e }{ 5 } } +c$
• C. $\cfrac { { 5 }^{ -x } }{ \log _{ e }{ 5 } } -\cfrac { 2^{ -x } }{ \log _{ e }{ 2 } } +c$
• D. $\cfrac { { 2 }^{ -x } }{ \log _{ e }{ 2 } } -\cfrac { 5^{ -x } }{ \log _{ e }{ 5 } } +c$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020