Evaluate the integral integrate of ((x-2)/(2x^2+7x+3))dx

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To evaluate the integral, we need to use partial fractions on the integrand.  This means that:

`{x-2}/{2x^2+7x+3}`

`={x-2}/{(2x+1)(x+3)}`

`=a/{2x+1}+b/{x+3}`   now use method of comparing coefficients

`x-2=a(x+3)+b(2x+1)`

The constant terms give

`-2=3a+b`   equation (1)

The linear terms give

`1=a+2b`   equation (2)

Now subtract (2)-2x(1)

`5=-5a`

so a=-1

Then we get b=1

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To evaluate the integral, we need to use partial fractions on the integrand.  This means that:

`{x-2}/{2x^2+7x+3}`

`={x-2}/{(2x+1)(x+3)}`

`=a/{2x+1}+b/{x+3}`   now use method of comparing coefficients

`x-2=a(x+3)+b(2x+1)`

The constant terms give

`-2=3a+b`   equation (1)

The linear terms give

`1=a+2b`   equation (2)

Now subtract (2)-2x(1)

`5=-5a`

so a=-1

Then we get b=1

This means the integral becomes:

`int{x-2}/{2x^2+7x+3}dx`

`=-int1/{2x+1}dx+int1/{x+3}dx`

`=-1/2ln(2x+1)+ln(x+3)+C`  where C is the constant of integration

`=ln({x+3}/(2x+1)^{1/2})+C`

The integral evaluates to `ln({x+3}/(2x+1)^{1/2})+C` .

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