How do you integrate #int ( 1/((x+1)^2+4)) dx# using partial fractions?

Question

Ezra Adams · Accepted Answer

#1/2arctan((x+1)/2)+C# #I=int1/((x+1)^2+4)dx# This isn't a job for partial fractions, it's a job for a trig substitution. Let #x+1=2tantheta#. This implies that #dx=2sec^2thetad theta#. Substituting in tells us: #I=int1/(4tan^2theta+4)(2sec^2thetad theta)=1/2intsec^2theta/(1+tan^2theta)d theta# Since #1+tan^2theta=sec^2theta#: #I=1/2intd theta=1/2theta+C# Undoing the substitution #x+1=2tantheta#: #I=1/2arctan((x+1)/2)+C#

Emily Ammerman · Answer

undefined To integrate $ \int \frac{1}{(x+1)^2 + 4} \, dx $ using partial fractions, follow these steps:

1. Recognize that the denominator $ (x+1)^2 + 4 $ cannot be factored further.
2. Express the fraction as a sum of partial fractions with undetermined coefficients: $ \frac{1}{(x+1)^2 + 4} = \frac{A}{x+1} + \frac{B}{(x+1)^2 + 4} $.
3. Clear the fractions by multiplying both sides of the equation by $ (x+1)^2 + 4 $.
4. Combine the terms to obtain a single fraction.
5. Equate the numerators of the fractions and solve for the unknown coefficients.
6. Once you've found the values of $ A $ and $ B $, integrate each partial fraction separately.
7. Add the integrated partial fractions together to get the final result.

Therefore, the integral of $ \frac{1}{(x+1)^2 + 4} $ using partial fractions results in the sum of two terms, one involving the natural logarithm and the other involving arctangent.

Aurora Alvarado · Answer

undefined To integrate $ \int \frac{1}{(x+1)^2+4} $ using partial fractions, first complete the square in the denominator:

$ (x+1)^2 + 4 = (x+1)^2 + 2^2 $

Now, let's rewrite the integral with a slightly different form:

$ \int \frac{1}{(x+1)^2+2^2} \, dx $

To decompose this into partial fractions, assume:

$ \frac{1}{(x+1)^2+2^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2+2^2} $

Now, multiply both sides by the denominator to clear the fraction:

$ 1 = A((x+1)^2+2^2) + B(x+1) $

Expand and simplify:

$ 1 = A(x^2 + 2x + 1 + 4) + B(x+1) $

$ 1 = Ax^2 + (2A + B)x + (5A + B) $

Now, equate coefficients of like terms:

For constants: $ 1 = 5A + B $  
For $ x $-terms: $ 0 = 2A + B $  
For $ x^2 $-terms: $ 0 = A $

Solve these simultaneous equations to find the values of $ A $ and $ B $.

$ A = 0 $  
$ B = -2 $

Now, rewrite the integral using the partial fractions:

$ \int \frac{1}{(x+1)^2+4} \, dx = \int \frac{-2}{(x+1)^2+4} \, dx $

The integral of $ \frac{-2}{(x+1)^2+4} $ can be evaluated by using a trigonometric substitution or other methods for integration of rational functions.