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

Question

Ezra Adams · Accepted Answer

#int(x-1)/(1+x^2)dx=1/2ln(1+x^2)-tan^(-1)x# As in the given algebraic fraction, we have the only denominator #1+x^2#, which cannot be factorized. As it is a quadratic function, in partial decomposition, the numerator would be in form #ax+b#. But it is already in this form. Hence we cannot convert them into further partial fractions. Now let #u=1+x^2#, hence #du=2xdx# Hence, #int(x-1)/(1+x^2)dx# = #intx/(1+x^2)dx-int1/(1+x^2)dx# Now #intx/(1+x^2)dx=int(du)/(2u)=1/2xxlnu=(ln(1+x^2))/2# and #int1/(1+x^2)dx=tan^(-1)x# (as if #tan^(-1)x=v#, #x=tanv# and #(dx)/(dv)=sec^2v=1+tan^2v=1+x^2#, hence #(dx)/(1+x^2)=dv# and #int(dx)/(1+x^2)=intdv=v=tan^(-1)x#) Hence #int(x-1)/(1+x^2)dx=1/2ln(1+x^2)-tan^(-1)x#

Paisley Johnson · Answer

undefined To integrate $ \frac{x - 1}{1 + x^2} $ using partial fractions, we first decompose the fraction into partial fractions as follows:

$$ \frac{x - 1}{1 + x^2} = \frac{A}{x^2 + 1} + \frac{Bx + C}{1 + x^2} $$

Multiplying both sides by $ 1 + x^2 $ to clear the denominators gives:

$$ x - 1 = A(x^2 + 1) + (Bx + C) $$

Expanding and combining like terms yields:

$$ x - 1 = Ax^2 + A + Bx + C $$

Now, equating coefficients of corresponding powers of x gives us the following system of equations:

$$ A = 0 $$
$$ B = 1 $$
$$ A + C = -1 $$

From the first equation, we get $ A = 0 $. Substituting this into the third equation gives $ C = -1 $. Therefore, $ B = 1 $.

The partial fraction decomposition is then:

$$ \frac{x - 1}{1 + x^2} = \frac{1}{1 + x^2} - \frac{1}{x^2 + 1} $$

Now, we can integrate each term separately:

$$ \int \frac{1}{1 + x^2} dx = \arctan(x) + C_1 $$
$$ \int \frac{1}{x^2 + 1} dx = \arctan(x) + C_2 $$

Where $ C_1 $ and $ C_2 $ are constants of integration. Thus, the integral of $ \frac{x - 1}{1 + x^2} $ is:

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

Where $ C $ is the constant of integration.

Evelyn Agard · Answer

undefined To integrate $ \frac{x - 1}{1 + x^2} $ using partial fractions, follow these steps:

1. Express the given rational function as a sum of partial fractions.
2. Integrate each partial fraction separately.
3. Combine the results to obtain the final integrated expression.

The partial fraction decomposition of $ \frac{x - 1}{1 + x^2} $ can be written as:

$$ \frac{x - 1}{1 + x^2} = \frac{A}{1 + x} + \frac{Bx + C}{1 + x^2} $$

where $ A $, $ B $, and $ C $ are constants to be determined.

After finding the values of $ A $, $ B $, and $ C $, integrate each partial fraction separately. Then, combine the results to obtain the final integrated expression.

The integral of $ \frac{A}{1 + x} $ is $ A \ln|1 + x| $, and the integral of $ \frac{Bx + C}{1 + x^2} $ can be computed using arctangent function.

Finally, combine the results to obtain the final integrated expression.