# How do you integrate #int x^3/[x(x^2+2x+1)]# using partial fractions?

Thus

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To integrate ( \frac{x^3}{x(x^2+2x+1)} ) using partial fractions, first factor the denominator: ( x(x^2+2x+1) = x(x+1)^2 ). Then, express the given fraction as the sum of two fractions with unknown numerators ( A ) and ( B ):

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

Clear the denominators by multiplying both sides by ( x(x+1)^2 ):

[ x^3 = A(x+1)^2 + Bx ]

Expand and collect like terms:

[ x^3 = A(x^2 + 2x + 1) + Bx ]

[ x^3 = Ax^2 + 2Ax + A + Bx ]

Now, equate the coefficients of like terms:

For the ( x^3 ) terms: ( 0x^3 = A )

For the ( x^2 ) terms: ( 1x^2 = A )

For the ( x ) terms: ( 0x = 2A + B )

For the constant terms: ( 0 = A )

Solving these equations:

[ A = 0 ] [ B = 0 ]

Now, we have ( A = 0 ) and ( B = 0 ), which means the partial fraction decomposition is ( \frac{0}{x} + \frac{0}{(x+1)^2} ).

Thus, the integral of ( \frac{x^3}{x(x^2+2x+1)} ) using partial fractions is simply ( \int \frac{x^3}{x(x^2+2x+1)} , dx = \int 0 , dx = C ), where ( C ) is the constant of integration.

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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.

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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