How do you find the antiderivative of #int 1/(x^2+3x) dx#?

Answer 1

#1/3ln|x|-1/3ln|x+3|+C#.

Write down: #=int1/(x(x+3))dx# #=int( { }/x + { }/(x+3))dx#.
Then apply the cover-up rule for partial fractions. To find out what goes over the #x#, use your finger to cover up the factor #x# in the denominator of the fraction on the first line, and replace all other #x#'s with zero. Similarly, to find out what goes over the #x+3#, cover up the #x+3# and replace the other #x# with #-3#. In each case, you replace #x# with whatever value of #x# makes the expression under your finger zero.
#=int( (1)/(0+3))/x + (1/(-3))/(x+3)dx# #=1/3int1/x-1/(x+3)dx#.
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Answer 2

To find the antiderivative of ( \frac{1}{x^2 + 3x} ) with respect to ( x ), we can use partial fraction decomposition:

[ \frac{1}{x^2 + 3x} = \frac{1}{x(x + 3)} ]

Now, we need to decompose ( \frac{1}{x(x + 3)} ) into partial fractions. We express ( \frac{1}{x(x + 3)} ) as the sum of two fractions with undetermined constants:

[ \frac{1}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3} ]

Now, we need to find the values of ( A ) and ( B ). We can do this by finding a common denominator and equating coefficients:

[ 1 = A(x + 3) + Bx ]

Solving for ( A ) and ( B ), we get:

[ A(x + 3) + Bx = Ax + 3A + Bx ] [ 1 = (A + B)x + 3A ]

For this equation to hold true for all ( x ), the coefficients of corresponding powers of ( x ) must be equal. Thus, we have two equations:

  1. Coefficient of ( x ): ( A + B = 0 )
  2. Constant term: ( 3A = 1 )

From the second equation, ( A = \frac{1}{3} ). Substituting ( A = \frac{1}{3} ) into the first equation, we find ( B = -\frac{1}{3} ).

Now that we have found ( A ) and ( B ), we can rewrite the original integral as:

[ \int \frac{1}{x^2 + 3x} , dx = \int \left( \frac{1}{3x} - \frac{1}{3(x + 3)} \right) , dx ]

[ = \frac{1}{3} \int \frac{1}{x} , dx - \frac{1}{3} \int \frac{1}{x + 3} , dx ]

Now, integrate each term:

[ = \frac{1}{3} \ln|x| - \frac{1}{3} \ln|x + 3| + C ]

Where ( C ) is the constant of integration.

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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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