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

Answer 1

#I= -1/4ln(x^2+3)+1/(2sqrt3)arctan(x/sqrt3) +1/2ln|x+3| +C#

#(3-x)/((x^2+3)(x+3))=(Ax+B)/(x^2+3)+C/(x+3)=#
#=((Ax+B)(x+3)+C(x^2+3))/((x^2+3)(x+3))=#
#=(Ax^2+Bx+3Ax+3B+Cx^2+3C)/((x^2+3)(x+3))=#
#=((A+C)x^2+(B+3A)x+(3B+3C))/((x^2+3)(x+3))#
#A+C=0 => A=-C# #B+3A=-1 => B=-1+3C# #3B+3C=3 => B+C=1#
#-1+3C+C=1 => 4C=2 => C=1/2#
#A=-1/2#
#B=1-C=1/2#
#I=int (3-x)/((x^2+3)(x+3))dx = int (-1/2x+1/2)/(x^2+3)dx + 1/2intdx/(x+3)#
#I=-1/4int(2xdx)/(x^2+3)+1/2intdx/(x^2+(sqrt3)^2)+1/2ln|x+3|#
#I= -1/4ln(x^2+3)+1/(2sqrt3)arctan(x/sqrt3) +1/2ln|x+3| +C#
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Answer 2

To integrate ( \frac{3 - x}{(x^2 + 3)(x + 3)} ) using partial fractions, first, we need to decompose the fraction into partial fractions. We start by expressing the fraction as:

[ \frac{3 - x}{(x^2 + 3)(x + 3)} = \frac{A}{x + 3} + \frac{Bx + C}{x^2 + 3} ]

Then, we find the values of ( A ), ( B ), and ( C ) by equating numerators:

[ 3 - x = A(x^2 + 3) + (Bx + C)(x + 3) ]

After expanding and collecting like terms, we can solve for ( A ), ( B ), and ( C ).

Next, we integrate each term of the partial fractions separately.

[ \int \frac{A}{x + 3} , dx + \int \frac{Bx + C}{x^2 + 3} , dx ]

For the first term, the integral is straightforward:

[ \int \frac{A}{x + 3} , dx = A \ln|x + 3| + C_1 ]

For the second term, we can use trigonometric substitution or complete the square to integrate.

Once we've integrated both terms, we combine the results to get the final answer.

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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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