Mathematics

Solve :
$$\int_{}^{} {\frac{{\left( {x - 1} \right)}}{{\left( {x + 1} \right)\left( {x - 2} \right)}}} dx$$


SOLUTION
$$\begin{array}{l} \int _{  }^{  }{ \frac { { \left( { x-1 } \right)  } }{ { \left( { x+1 } \right) \left( { x-2 } \right)  } }  } dx \\ Let\, \frac { { \left( { x-1 } \right)  } }{ { \left( { x+1 } \right) \left( { x-2 } \right)  } } =\frac { A }{ { \left( { x+1 } \right)  } } +\frac { B }{ { \left( { x-2 } \right)  } }  \\ Then,\left( { x-1 } \right) \equiv A\left( { x-2 } \right) +B\left( { x+1 } \right) ......\left( i \right)  \\ putting\, x=1\, in\left( i \right) ,\, we\, get\, A=\frac { 2 }{ 3 } . \\ putting\, x=2\, in\left( i \right) ,\, we\, get\, B=\frac { 1 }{ 3 } . \\ \therefore \frac { { \left( { x-1 } \right)  } }{ { \left( { x+1 } \right) \left( { x-2 } \right)  } } =\frac { 2 }{ { 3\left( { x+1 } \right)  } } +\frac { 1 }{ { 3\left( { x-2 } \right)  } }  \\ =\int _{  }^{  }{ \frac { { \left( { x-1 } \right)  } }{ { \left( { x+1 } \right) \left( { x-2 } \right)  } }  } dx \\ =\frac { 2 }{ 3 } \int _{  }^{  }{ \frac { { dx } }{ { \left( { x+1 } \right)  } }  } +\int _{  }^{  }{ \frac { { dx } }{ { \left( { x-2 } \right)  } }  }  \\ =\frac { 2 }{ 3 } \log  \left| { x+1 } \right| +\frac { 1 }{ 3 } \log  \left| { x-2 } \right| +C \end{array}$$
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 84
Enroll Now For FREE

Realted Questions

Q1 Single Correct Hard
$$\displaystyle \int \sqrt {1+x \sqrt {1+(x+1) \sqrt {1+(x+2) (x+4)}}}$$ $$dx$$ is equal to
  • A. $$\displaystyle \frac{x^2}{2} - x+ c$$
  • B. $$\displaystyle \frac{x^2}{2} + c$$
  • C. $$x+c$$
  • D. $$\displaystyle \frac{x^2}{2} + x+ c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
$$I = \displaystyle \int{\dfrac{x\cos x+\sin x}{x\sin x}}dx$$ Then $$I = ?$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Medium
Evaluate $$\displaystyle\int {\frac{{2t\,dt}}{{3t\, + 4}}} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Multiple Correct Hard
If $$\displaystyle I=\int \frac{dx}{\sqrt{2x+3}+\sqrt{x+2}},$$ then $$I$$ equals

  • A. $$\displaystyle \log \left | \frac{\sqrt{x+2}+\sqrt{2x+3}}{\sqrt{x+2}-\sqrt{2x+3}} \right |+C$$
  • B. $$\displaystyle \left ( \sqrt{x+2}+\sqrt{2x+3} \right )+C$$
  • C. $$\displaystyle 2\left ( u-v \right )+\log\left | \frac{u-1}{u+1} \right |+\log\left | \frac{v-1}{v+1} \right |+C, u=\sqrt{2x+3},v=\sqrt{x+2}$$
  • D. is transcedental function in u and v, $$\displaystyle u=\sqrt{2x+3}v=\sqrt{x+2}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Passage Medium
The average value of a function f(x) over the interval, [a,b] is the number $$\displaystyle \mu =\frac{1}{b-a}\int_{a}^{b}f\left ( x \right )dx$$
The square root $$\displaystyle \left \{ \frac{1}{b-a}\int_{a}^{b}\left [ f\left ( x \right ) \right ]^{2}dx \right \}^{1/2}$$ is called the root mean square of f on [a, b]. The average value of $$\displaystyle \mu $$ is attained id f is continuous on [a, b].

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer