Mathematics

# Solve:$\int { \dfrac { dx }{ 2{ x }^{ 2 }+x-1 } }$

##### SOLUTION

We have,

$\int{\dfrac{dx}{2{{x}^{2}}+x-1}}$

$\Rightarrow \int{\dfrac{dx}{2{{x}^{2}}+\left( 2-1 \right)x-1}}$

$\Rightarrow \int{\dfrac{dx}{2{{x}^{2}}+2x-1x-1}}$

$\Rightarrow \int{\dfrac{dx}{2x\left( x+1 \right)-1\left( x+1 \right)}}$

$\Rightarrow \int{\dfrac{dx}{\left( 2x-1 \right)\left( x+1 \right)}}$

Applying application of partial dfraction

$\dfrac{1}{\left( 2x-1 \right)\left( x+1 \right)}=\dfrac{A}{\left( 2x-1 \right)}+\dfrac{B}{\left( x+1 \right)}$

$1=A\left( x+1 \right)+B\left( 2x-1 \right)$

$1=Ax+A+2Bx-B$

$1=Ax+2Bx+A-B$

$1=x\left( A+2B \right)+\left( A-B \right)$

On comparing coefficients and we get,

$A+2B=0$

$A-B=1$

Solving both and we get,

$A=\dfrac{2}{3}$ and $B=\dfrac{-1}{3}$

Now,

$\dfrac{1}{\left( 2x-1 \right)\left( x+1 \right)}=\dfrac{2}{3\left( 2x-1 \right)}-\dfrac{1}{3\left( x+1 \right)}$

On integration and we get,

$\int{\dfrac{1}{\left( 2x-1 \right)\left( x+1 \right)}dx}=\int{\dfrac{2}{3\left( 2x-1 \right)}dx}-\int{\dfrac{1}{3\left( x+1 \right)}dx}$

$=\dfrac{2}{3}\dfrac{\log \left( 2x-1 \right)}{2}-\dfrac{1}{3}\log \left( x+1 \right)$

$=\dfrac{1}{3}\log \left( 2x-1 \right)-\dfrac{1}{3}\log \left( x+1 \right)$

$=\dfrac{1}{3}\log \dfrac{\left( 2x-1 \right)}{\left( x+1 \right)}$

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

#### Realted Questions

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$\displaystyle \overset{e^2}{\underset{1}{\int}} [log_e \,x]dx, x > 0$ and $[\cdot]$ is greatest integer function, is equal to
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1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
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Q3 Subjective Medium
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Q4 Subjective Medium
Evaluate the following integral as the limit of sum:
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