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

Question

Paisley Johnson · Accepted Answer

#1/2arctan((x+1)/2)+C#

Partial fractions cannot be used to express this, but we can integrate it with trigonometric substitutions.

#intdx/((x+1)^2+4)#

Let #x+1=2tantheta#. Thus, #dx=2sec^2thetad theta#. Substituting, this gives us:

#=int(2sec^2thetad theta)/(4tan^2theta+4)#

Factoring:

#=1/2int(sec^2thetad theta)/(tan^2theta+1)#

Recall that #tan^2theta+1=sec^2theta#:

#=1/2int(sec^2thetad theta)/sec^2theta#

#=1/2intd theta#

#=1/2theta+C#

From #x+1=2tantheta# we see that #theta=arctan((x+1)/2)#:

#=1/2arctan((x+1)/2)+C#

William Abdalla · Answer

undefined To integrate $ \frac{1}{(x+1)^2+4} $ using partial fractions, first express the integrand as a sum of simpler fractions:

$$ \frac{1}{(x+1)^2+4} = \frac{A}{x+1} + \frac{B}{(x+1)^2+2^2} $$

Then, find the values of A and B by equating the numerators of the fractions:

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

Solve for A and B by substituting appropriate values for x, typically those that will eliminate one of the fractions:

$$ x = -1 \rightarrow 1 = 2B \rightarrow B = \frac{1}{2} $$

$$ x = 0 \rightarrow 1 = A(2^2) + \frac{1}{2} \rightarrow A = \frac{3}{4} $$

Thus, the integral becomes:

$$ \int \left( \frac{3/4}{x+1} + \frac{1/2}{(x+1)^2+2^2} \right) dx $$

$$ = \frac{3}{4} \ln|x+1| + \frac{1}{2} \arctan \left( \frac{x+1}{2} \right) + C $$