Mathematics

Evaluate the following definite integral :
$$\displaystyle \int_{0}^{2}\dfrac {1}{4+x-x^2}dx$$


SOLUTION
$$I=\displaystyle \int_0^2 \dfrac {1}{4+x-x^2}dx=-\displaystyle \int_0^2 \dfrac {1}{x^2 -x-4}dx =-\displaystyle \int_0^2\dfrac {1}{\left (x-\dfrac {1}{2}\right)^2-\left (\dfrac {\sqrt {17}}{2}\right)}dx$$

$$\Rightarrow \ I=\displaystyle \int_0^2 \dfrac {1}{\left (\dfrac {\sqrt {17}}{2}\right)-\left (x-\dfrac {1}{2}\right)^2}dx =\dfrac {1}{\sqrt {17}} \left [\log \left (\dfrac {\sqrt {17}+2x-1}{\sqrt {17} -2x+1}\right) \right]_0^2$$

$$\Rightarrow \ I=\dfrac {1}{\sqrt {17}}\left\{\dfrac {\sqrt {17}+3}{\sqrt {17}-3} -\log \dfrac {\sqrt {17}-1}{\sqrt {17}+1} \right\}=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {52+12\sqrt {17}}{18-2\sqrt {17}} \times \dfrac {18+2\sqrt {17}}{18+2\sqrt {17}}\right)$$

$$\Rightarrow \ I=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {1344+320\sqrt {17}}{256}\right)=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {21+5\sqrt {17}}{4}\right)$$

View Full Answer

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86
Enroll Now For FREE

Realted Questions

Q1 Single Correct Medium
The function $$\displaystyle F(x)=\int_{0}^{x}log\frac{(1-x)}{1+x}dx$$ is a function that is 
  • A. Odd
  • B. Periodic
  • C. none of these
  • D. Even

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Hard
Evaluate $$\int\limits_0^{\frac{\pi }{4}} {{{\cos }^{\frac{3}{2}}}\left( {2x} \right)cos\left( x \right)dx} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Medium
If $$\int { \cfrac { \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+4 \right)  }{ \left( { x }^{ 2 }+3 \right) \left( { x }^{ 2 }-5 \right)  }  } dx=\int { \left\{ 1+\cfrac { f(x) }{ \left( { x }^{ 2 }+3 \right) \left( { x }^{ 2 }-5 \right)  }  \right\}  } dx$$
$$x+A\tan ^{ -1 }{ \left( \cfrac { x }{ A' }  \right)  } +B\log { \left( \cfrac { x-l }{ x+m }  \right)  } +K\quad $$ then which of the following is correct
  • A. $$A=\cfrac { 1 }{ 4\sqrt { 3 } } ,B=\cfrac { 27 }{ 8\sqrt { 5 } } ,K\in R$$
  • B. $$f(x)=7{ x }^{ 2 }+19,A'=\sqrt { 3 } ,K\in R$$
  • C. $$l=m=\sqrt { 5 } ,L=1,K\in R$$
  • D. All of these

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Single Correct Hard
Number of positive continuous functions $$f(x)$$ defined in $$[0,1]$$ for which $$\displaystyle\int_0^1{f(x)dx}=1$$, $$\displaystyle\int_0^1{xf(x)dx}=2$$, $$\displaystyle\int_0^1{x^2f(x)dx}=4$$, is
  • A. $$1$$
  • B. $$4$$
  • C. Infinite
  • D. None of these

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Single Correct Medium
$$ 4 \displaystyle \int \dfrac{\sqrt{a^6 + x^8}}{x} dx$$ is equal to ______________________.
  • A. $$a^6 ln\vert \dfrac {\sqrt{a^6 + x^8} - a^3} {\sqrt {a^6 + x^8} + a^3}\vert + c$$
  • B. $$\sqrt{a^6 + x^8} + \dfrac {a^3}{2} ln \vert \dfrac {\sqrt{a^6 + x^8} - a^3} {\sqrt {a^6 + x^8} + a^3}\vert + c$$
  • C. $$a^6 ln \vert \dfrac {\sqrt{a^6 + x^8} + a^3} {\sqrt {a^6 + x^8} - a^3}\vert + c$$
  • D. $$\sqrt{a^6 + x^8} + \dfrac{a^3}{2} ln \vert \dfrac {\sqrt{a^6 + x^8} + a^3} {\sqrt {a^6 + x^8} - a^3}\vert + c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer