Let f(x)= 3x^3+6x-8÷x(x^2+2) (i) Express f(x) in the form (A)+(B÷x)+(Cx+D÷x^2+2). (ii) show the integral 2 to 1 f(x) dx=3-ln4 Can someone please solve this question?

Answer 1

Please see below.

As #f(x)=(3x^3+6x-8)/(x(x^2+2))=(3x^3+6x-8)/(x^3+2x)#
= #3-8/(x^3+2x)=3-8/(x(x^2+2x)#
Here we have divided #3x^3+6x-8# by #x^2+2x#, which leads to #A=3#
Let #8/(x(x^2+2x))=B/x+(Cx+D)/(x^2+2)#
then #8=B(x^2+2)+x(Cx+D)#

Comparing coeeficients of like powers, we get

#2B=8# i.e. #B=4# #B+C=0# i.e. #C=-4# #D=0#
Hence #f(x)=(3x^3+6x-8)/(x(x^2+2))=3-4/x+(4x)/(x^2+2)#
and #I=int_1^2(3x^3+6x-8)/(x(x^2+2))dx#
= #int_1^2(3-4/x+(4x)/(x^2+2))dx#
= #3int_1^2dx-4int_1^2(dx)/x+2int_1^2(2x)/(x^2+2)dx#
= #[3x-4lnx]_1^2+2int_1^2(2x)/(x^2+2)dx#
For #int_1^2(2x)/(x^2+2)dx#, let #u=x^2+2#, then #du=2xdx#
and #int_1^2(2x)/(x^2+2)dx=int_3^6(du)/u=[ln6-ln3]#
Hence #I=[3x-4lnx]_1^2+2[ln6-ln3]#
= #6-4ln2-3+2ln2#
= #3-2ln2#
= #3-ln4#
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Answer 2

(i) ( f(x) = \frac{3x^3 + 6x - 8}{x(x^2 + 2)} )

To express ( f(x) ) in the form ( A + \frac{B}{x} + \frac{Cx + D}{x^2 + 2} ), first decompose the rational function into partial fractions:

( f(x) = \frac{3x^3 + 6x - 8}{x(x^2 + 2)} = A + \frac{B}{x} + \frac{Cx + D}{x^2 + 2} )

Multiplying both sides by ( x(x^2 + 2) ) to clear the denominators:

( 3x^3 + 6x - 8 = A(x^2 + 2) + B(x^2 + 2) + (Cx + D)x )

( 3x^3 + 6x - 8 = Ax^2 + 2A + Bx^2 + 2B + Cx^2 + Dx )

Now, equate coefficients of like terms:

( 3x^3 + 6x - 8 = (A + B)x^2 + (D + C)x + 2A + 2B )

Comparing coefficients:

( A + B = 3 ) (coefficient of ( x^2 )) ( D + C = 6 ) (coefficient of ( x )) ( 2A + 2B = -8 ) (constant term)

Now, solve this system of equations to find the values of ( A, B, C, ) and ( D ).

(ii)

[ \int_{1}^{2} f(x) , dx = 3 - \ln(4) ]

Given ( f(x) ) in the form ( A + \frac{B}{x} + \frac{Cx + D}{x^2 + 2} ), integrate each term separately within the given limits and then evaluate the definite integral.

[ \int_{1}^{2} f(x) , dx = \int_{1}^{2} A , dx + \int_{1}^{2} \frac{B}{x} , dx + \int_{1}^{2} \frac{Cx + D}{x^2 + 2} , dx ]

[ = A(x)\bigg|{1}^{2} + B\ln|x|\bigg|{1}^{2} + \left(\frac{C}{2}\ln(x^2 + 2) + \frac{D}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt{2}}\right)\right)\bigg|_{1}^{2} ]

[ = A(2) - A(1) + B\ln(2) - B\ln(1) + \left(\frac{C}{2}\ln(6) + \frac{D}{\sqrt{2}}\arctan\left(\frac{2}{\sqrt{2}}\right)\right) - \left(\frac{C}{2}\ln(3) + \frac{D}{\sqrt{2}}\arctan\left(\frac{1}{\sqrt{2}}\right)\right) ]

[ = A + B\ln(2) - B + \frac{C}{2}\ln(6) + \frac{D}{\sqrt{2}}\left(\arctan\left(\frac{2}{\sqrt{2}}\right) - \arctan\left(\frac{1}{\sqrt{2}}\right)\right) - \frac{C}{2}\ln(3) ]

Given that ( \int_{1}^{2} f(x) , dx = 3 - \ln(4) ), equate the expressions:

[ A + B\ln(2) - B + \frac{C}{2}\ln(6) + \frac{D}{\sqrt{2}}\left(\arctan\left(\frac{2}{\sqrt{2}}\right) - \arctan\left(\frac{1}{\sqrt{2}}\right)\right) - \frac{C}{2}\ln(3) = 3 - \ln(4) ]

Now, substitute the values of ( A, B, C, ) and ( D ) obtained from part (i) into this equation and solve for any remaining unknowns to verify the given integral.

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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.

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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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