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

Question

Scarlett Allison · Accepted Answer

#I=1/2ln|x^2-2|+1/(2sqrt2)ln|(x-sqrt2)/(x+sqrt2)|+c#

We know that,

#color(red)((1)int(d/(dx)(f(x)))/(f(x))dx=ln|f(x)|+c#

#color(blue)((2)int1/(x^2-a^2)dx=1/(2a)ln|(x-a)/(x+a)|+c#

Here,

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

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

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

#=int[x/(x^2-2)+1/(x^2-2)]dx#

#=1/2int(2x)/(x^2-2)dx+int1/(x^2-(sqrt2)^2)dx#

Using #color(red)((1)) and color(blue)((2)) #, we get

#I=1/2ln|x^2-2|+1/(2sqrt2)ln|(x-sqrt2)/(x+sqrt2)|+c#

Evelyn Agard · Answer

Given: #int (x^2+2x+1)/((x+1)(x^2-2)) dx#

The numerator factors:

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

Cancel the common factor:

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

The denominator can be factored as the difference of two squares:

#int (x+1)/((x-sqrt2)(x+sqrt2) dx#

Write the partial fractions equation:

#(x+1)/((x-sqrt2)(x+sqrt2)) = A/(x-sqrt2)+B/(x+sqrt2)#

Multiply both sides by #(x-sqrt2)(x+sqrt2)#:

#x+1 = A(x+sqrt2)+B(x-sqrt2)#

Let #x = sqrt2#:

#sqrt2+1 = A(sqrt2+sqrt2)+B(sqrt2-sqrt2)#

#sqrt2+1 = A(2sqrt2)#

#A = (2+sqrt2)/4#

Let #x = -sqrt2#:

#-sqrt2+1 = A(-sqrt2+sqrt2)+B(-sqrt2-sqrt2)#

#-sqrt2+1 = B(-2sqrt2)#

#B = (2-sqrt2)/4#

Write in integral form:

#(2+sqrt2)/4 int 1/(x-sqrt2) dx + (2-sqrt2)/4 int 1/(x+sqrt2) dx#

The integrals become natural logarithms:

#(2+sqrt2)/4 ln(x-sqrt2) + (2-sqrt2)/4 ln(x+sqrt2)+ C#

Scarlett Allison · Answer

undefined To find the integral of $\frac{x^2+2x+1}{(x+1)(x^2-2)}$ using partial fractions, first factor the denominator completely: $(x+1)(x^2-2)$.

The denominator can be factored as $ (x+1)(x-\sqrt{2})(x+\sqrt{2}) $.

The partial fraction decomposition will have the form:

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

Multiply both sides by $(x+1)(x^2-2)$ to clear the fractions:

$x^2 + 2x + 1 = A(x^2 - 2) + B(x+1)(x+\sqrt{2}) + C(x+1)(x-\sqrt{2})$

Now, solve for A, B, and C by comparing coefficients.

1. Substitute $x = -1$: $A = \frac{1}{3}$.
2. Substitute $x = \sqrt{2}$: $B = \frac{1}{3\sqrt{2}}$.
3. Substitute $x = -\sqrt{2}$: $C = \frac{-1}{3\sqrt{2}}$.

Now, rewrite the integral using the partial fraction decomposition and integrate term by term:

$$\int \left( \frac{1}{3(x+1)} + \frac{1}{3\sqrt{2}(x-\sqrt{2})} - \frac{1}{3\sqrt{2}(x+\sqrt{2})} \right) \, dx$$

Integrate each term separately:

$$= \frac{1}{3} \ln|x+1| + \frac{1}{3\sqrt{2}} \ln|x-\sqrt{2}| - \frac{1}{3\sqrt{2}} \ln|x+\sqrt{2}| + C$$

Where $C$ is the constant of integration.