How do you integrate #int (x+4)/(x^2 + 2x + 5)# using partial fractions?

Answer 1

# int \ (x+4)/(x^2+2x+5) \ dx = 1/2 \ ln |x^2+2x+5| + 3/2 \ arctan((x+1)/2) + C #

We seek:

# I = int \ (x+4)/(x^2+2x+5) \ dx #

Since we know that the denominator does not factorize with real factors, we can decompose the integrand as follows instead of using partial fractions:

# I = int \ (1/2(2x+2) +3)/(x^2+2x+5) \ dx #

When the numerator has been altered to become the denominator's derivative, this means that:

# I = int \ (1/2(2x+2))/(x^2+2x+5) +3/(x^2+2x+5) \ dx # # \ \ = 1/2 \ int \ (2x+2)/(x^2+2x+5) \ dx +3 \ int 1/(x^2+2x+5) \ dx #
For the First Integral, #I_1#, say, we can use a substitution, Lte
# u = x^2+2x+5 => (du)/dx = 2x+2 #

After that, substituting results in (without the integration constant):

# I_1 = 1/2 \ int \ 1/u \ du # # \ \ \ = 1/2 \ ln|u| #

After making the substitution again, we obtain:

# I_1 = 1/2 \ ln |x^2+2x+5| #
Next, we consider the second integral, #I_2#, say, we complete the square of the denominator, thus:
# I_2 = 3 \ int 1/((x+1)^2-1^2+5) \ dx # # \ \ \ = 3 \ int 1/((x+1)^2+4) \ dx #

Additionally, we are able to substitute, Let:

# u = (x+1)/2 => (du)/dx = 1/2 #

And by substituting, we obtain (without the integration constant):

# I_2 = 3 \ int 2/((2u)^2+4) \ du # # \ \ \ = 3 \ int 2/(4u^2+4) \ du # # \ \ \ = 3/2 \ int 1/(u^2+1) \ du # # \ \ \ = 3/2 \ arctan u #

After making the substitution again, we obtain:

# I_2 = 3/2 \ arctan((x+1)/2) #

After combining the two results and adding the integration constant, we obtain:

# I = 1/2 \ ln |x^2+2x+5| + 3/2 \ arctan((x+1)/2) + C #
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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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