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,

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

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

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

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

Hence, this is the answer.
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