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

Answer 1

The answer is #=ln(x+1)-1/2ln(x^2+2x+2)+C#

The decomposition in partial fractions will be discussed first.

#1/((x+1)(x^2+2x+2))=A/(x+1)+(Bx+C)/(x^2+2x+2)#
#=(A(x^2+2x+2)+(Bx+C)(x+1))/((x+1)(x^2+2x+2))#
So, #1=A(x^2+2x+2)+(Bx+C)(x+1)#
Let #x=0#, #=>#, #1=2A+C#
Let #x=-1#, #=>#, #1=A#
coefficients of x^2, #0=A+B# #=>#, #B=-1#

#C=-1

#1/((x+1)(x^2+2x+2))=1/(x+1)+(-x-1)/(x^2+2x+2)#
#=1/(x+1)-(x+1)/(x^2+2x+2)#
#int(dx)/((x+1)(x^2+2x+2))=intdx/(x+1)-int(((x+1))dx)/(x^2+2x+2)#
#intdx/(x+1)=ln(x+1)#

In order to get the second integral, we substitute

#u=x^2+2x+2#
#du=(2x+2)dx=2(x+1)dx#
#int(((x+1))dx)/(x^2+2x+2)=1/2int(du)/u=1/2lnu#
#=1/2ln(x^2+2x+2)#

When everything is combined,

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

To integrate ( \frac{1}{(x+1)(x^2 + 2x + 2)} ) using partial fractions, follow these steps:

  1. Decompose the fraction into partial fractions: ( \frac{1}{(x+1)(x^2 + 2x + 2)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 2x + 2} )

  2. Clear the denominator by multiplying both sides by ( (x+1)(x^2 + 2x + 2) ): ( 1 = A(x^2 + 2x + 2) + (Bx + C)(x + 1) )

  3. Expand and group like terms: ( 1 = Ax^2 + 2Ax + 2A + Bx^2 + Bx + Cx + C )

  4. Equate coefficients of like terms: ( A + B = 0 ) (for ( x^2 ) terms)
    ( 2A + B + C = 0 ) (for ( x ) terms)
    ( 2A + C = 1 ) (for constant terms)

  5. Solve the system of equations to find the values of ( A ), ( B ), and ( C ).

  6. Once you have found the values of ( A ), ( B ), and ( C ), rewrite the original fraction with the partial fractions:

    ( \frac{1}{(x+1)(x^2 + 2x + 2)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 2x + 2} )

  7. Now integrate each partial fraction separately.

    ( \int \frac{1}{(x+1)(x^2 + 2x + 2)} , dx = \int \frac{A}{x+1} , dx + \int \frac{Bx + C}{x^2 + 2x + 2} , dx )

    ( = A \ln|x+1| + \int \frac{Bx + C}{x^2 + 2x + 2} , dx )

  8. For the second integral, complete the square in the denominator, then use a substitution method to integrate.

  9. After integrating both parts, you'll have your final result.

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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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