How do you find #int (x^2 + 1)/ (2x^3 - 5x^2 + x + 2)dx# using partial fractions?

Answer 1

The answer is #=-2/3ln(∣x-1∣)+ln(∣x-2∣)+1/6ln(∣2x+1∣)+C#

Let #f(x)=2x^3-5x^2+x+2#
#f(1)=2-5+1+2=0#
Therefore, #(x-1)# is a factor
#f(2)=16-20+2+2=0#
Therefore, #(x-2)# is a factor
#(x-1)(x-2)=x^2-3x+2# is a factor

To find the last factor, we do a long division

#color(white)(aaaa)##2x^3-5x^2+x+2##color(white)(aaaa)##∣##x^2-3x+2#
#color(white)(aaaa)##2x^3-6x^2+4x##color(white)(aaaaaaa)##∣##2x+1#
#color(white)(aaaaaa)##0+x^2-3x+2#
#color(white)(aaaaaaaa)##+x^2-3x+2#
#color(white)(aaaaaaaaaa)##+0-0+0#

We can perform the decomposition into partial fractions

#(x^2+1)/(2x^3-5x^2+x+2)=(x^2+1)/((x-1)(x-2)(2x+1))#
#=A/(x-1)+B/(x-2)+C/(2x+1)#
#=(A(x-2)(2x+1)+B(x-1)(2x+1)+C(x-1)(x-2))/((x-1)(x-2)(2x+1))#

Therefore,

#x^2+1=A(x-2)(2x+1)+B(x-1)(2x+1)+C(x-1)(x-2))#
Let #x=1#, #=>#, #2=-3A#, #=>#, #A=-2/3#
Let #x=2#, #=>#, #5=5B#, #=>#, #B=1#
Let #x=-1/2#, #=>#, #5/4=15/4C#, #=>#, #C=1/3#

So,

#(x^2+1)/(2x^3-5x^2+x+2)=(-2/3)/(x-1)+1/(x-2)+(1/3)/(2x+1)#
#int((x^2+1)dx)/(2x^3-5x^2+x+2)=int((-2/3)dx)/(x-1)+int(dx)/(x-2)+int((1/3)dx)/(2x+1)#
#=-2/3ln(∣x-1∣)+ln(∣x-2∣)+1/6ln(∣2x+1∣)+C#
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Answer 2

To find ∫(x^2 + 1)/(2x^3 - 5x^2 + x + 2) dx using partial fractions, first factorize the denominator. Then, express the fraction as a sum of simpler fractions using partial fractions decomposition. You'll have three linear factors in the denominator. Set up the partial fraction decomposition with undetermined coefficients. After finding the coefficients, integrate each term separately.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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