How do you integrate $\int \frac{2 x + 1}{(x - 2) (x^{2} + 4)}$ $int (2x+1) /( (x-2)(x^2+4))$ using partial fractions?

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How do you integrate $\int \frac{2 x + 1}{(x - 2) (x^{2} + 4)}$ $int (2x+1) /( (x-2)(x^2+4))$ using partial fractions?

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Answer 1 · 2017-09-30T07:53:10

$\frac{(5 - ln | x - 2 |) - (3 \cdot {tan}^{- 1} (\frac{x}{2}))}{8} + \frac{5 ln ∣ ∣ x^{2} + 4 ∣ ∣}{16} + c$ $((5-ln|x-2|)-(3*tan^-1(x/2)))/8+(5ln|x^2+4|)/16+c$

Explanation:

First step in this problem is understanding that the use of partial fractions is valid since the numerator is a lesser degree of power than the denominator.
Apply Partial Fractions:
*Note that the denominator terms are Linear (left) and Quadratic (on the right), all this means is that they are terms that cannot be factorized further.
Linear form = $\frac{A}{x - 2}$ $A/(x-2)$
Quadratic form = $\frac{B x + C}{x^{2} + 4}$ $(Bx+C)/(x^2+4)$

Add these two terms together and set it equal to the Numerator.
$2 x + 1 = \frac{A}{x - 2} + \frac{B x + C}{x^{2} + 4}$ $2x+1=A/(x-2)+(Bx+C)/(x^2+4)$

Simplify by multiplying A by the denomenator of the other fraction, and the same for B:
$2 x + 1 = A (x^{2} + 4) + B x + C (x - 2)$ $2x+1=A(x^2+4)+Bx+C(x-2)$

now distribute:
$2 x + 1 = A x^{2} + 4 A + B x^{2} - 2 B x + C x - 2 C$ $2x+1=Ax^2+4A+Bx^2-2Bx+Cx-2C$

*Note: The biggest thing students confuse about partial fractions is this next step, in the left hand side you should think of it as $0 x^{2} + 2 x + 1$ $0x^2+2x+1$ to understand that there are Zero $x^{2}$ $x^2$ 's on the left.

$0 x^{2} + 2 x + 1 = A x^{2} + 4 A + B x^{2} - 2 B x + C x - 2 C$ $color(red)(0x^2)+color(blue)(2x)+color(green)(1)=color(red)(Ax^2)+color(green)(4A)+color(red)(Bx^2)-color(blue)(2Bx)+color(blue)(Cx)-color(green)(2C)$

Collect like terms:
$x^{2} \Rightarrow$ $color(red)(x^2) rArr$ $0 = A + B$ $0=A+B$
$x \Rightarrow$ $color(blue)x rArr$ $2 = - 2 B + C$ $2=-2B+C$
$Integers$ $color(green)"Integers"$ $\Rightarrow$ $rArr$ $1 = 4 A - 2 C$ $1=4A-2C$

Solve by any method, here we will use substitution
Eq1: $0 = A + B \Rightarrow [A = - B] \Leftrightarrow [B = - A]$ $0=A+B rArr [A=-B] hArr [B=-A]$
Eq2: $2 = - 2 B + C \Rightarrow [2 + 2 B = C]$ $2=-2B+C rArr [2+2B=C]$
Eq3: $1 = 4 A - 2 C \Rightarrow 1 = 2 (2 A - C) \Rightarrow [\frac{1}{2} = 2 A - C]$ $1=4A-2C rArr 1=2(2A-C) rArr [1/2=2A-C]$

Substitute in Eq1 and Eq2 into Eq3 to get:
$\frac{1}{2} = 2 (- B) - (2 + 2 B)$ $1/2 = 2(-B)-(2+2B)$
Simplify:
$\frac{1}{2} = - 4 B - 2$ $1/2=-4B-2$

Solve for B:
$B = - \frac{5}{8}$ $B=-5/8$
Use B to solve for A
$A = \frac{5}{8}$ $A=5/8$
Use B to solve for C
$2 + 2 (- \frac{5}{8}) = C$ $2+2(-5/8)=C$
$C = \frac{3}{4}$ $C=3/4$

Plug in A, B, and C values into original fraction:
$\frac{A}{x - 2} + \frac{B x + C}{x^{2} + 4}$ $A/(x-2)+(Bx+C)/(x^2+4)$

$\frac{\frac{5}{8}}{x - 2} + \frac{- \frac{5}{8} x + \frac{3}{4}}{x^{2} + 4}$ $(5/8)/(x-2)+(-5/8x+3/4)/(x^2+4)$

Simplify into:
$\frac{5}{8 (x - 2)} + \frac{- 5 x + 6}{8 (x^{2} + 4)}$ $5/(8(x-2))+(-5x+6)/(8(x^2+4))$

New integral is now represented by:
$\int \frac{5}{8 (x - 2)} d x + \int \frac{- 5 x + 6}{8 (x^{2} + 4)} d x$ $int5/(8(x-2))dx+int(-5x+6)/(8(x^2+4))dx$

Remove constants to prepare for integration:
$5 \cdot \int \frac{1}{8 (x - 2)} d x + \frac{1}{8} \cdot \int \frac{- 5 x + 6}{(x^{2} + 4)} d x$ $5*int1/(8(x-2))dx+1/8*int(-5x+6)/((x^2+4))dx$

Break apart second integral (on the right) into two separate integrals (Mind the 1/8 multiplier) and remove the constants in the numerators

$\frac{5}{8} \cdot \int \frac{1}{(x - 2)} d x + \frac{1}{8} \cdot (- 5 \cdot \int \frac{x}{x^{2} + 4} d x + 6 \cdot \int \frac{1}{x^{2} + 4} d x)$ $5/8*int1/((x-2))dx+1/8*(-5*intx/(x^2+4)dx+6*int1/(x^2+4)dx)$

First integral:
$\frac{5}{8} \int \frac{1}{x - 2} d x$ $5/8int1/(x-2)dx$
$u = x - 2$ $u=x-2$
Result is: $\frac{5 ln | x - 2 |}{8}$ $(5ln|x-2|)/8$

Second Integral:
$5 \cdot \int \frac{x}{x^{2} + 4} d x$ $5*intx/(x^2+4)dx$
$u = x^{2} + 4$ $u=x^2+4$
Result Is: $\frac{- 5 ln ∣ ∣ x^{2} + 4 ∣ ∣}{2}$ $(-5ln|x^2+4|)/2$

Third Integral:
$6 \cdot \int \frac{1}{x^{2} + 4} d x$ $6*int1/(x^2+4)dx$
Apply inverse trig formula by identifying denominator as Arctan:
$\int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \cdot {tan}^{- 1} (\frac{x}{a})$ $int1/(x^2+a^2)dx = 1/a * tan^-1(x/a)$
Where $[a^{2} = 4] \Rightarrow [a = 2]$ $[a^2=4] rArr [a=2]$
Result is: $3 {tan}^{- 1} (\frac{x}{2})$ $3tan^-1(x/2)$

Plug these back into full equation and do not forget the $\frac{1}{8}$ $1/8$ multiplier for the second and third integrals.

$\frac{5 \cdot ln | x - 2 |}{8} - \frac{1}{8} (\frac{- 5 ln ∣ ∣ x^{2} + 4 ∣ ∣}{2} + 3 {tan}^{- 1} (\frac{x}{2}))$ $(5*ln|x-2|)/8-1/8((-5ln|x^2+4|)/2+3tan^-1(x/2))$

Distribute the $\frac{1}{8}$ $1/8$ multiplier

$\frac{5 \cdot ln | x - 2 |}{8} + \frac{5 ln | x - 2 |}{16} - \frac{3 {tan}^{- 1} (\frac{x}{2})}{8} + c$ $(5*ln|x-2|)/8+(5ln|x-2|)/16-(3tan^-1(x/2))/8+c$

Simplify:
$\frac{(5 - ln | x - 2 |) - (3 \cdot {tan}^{- 1} (\frac{x}{2}))}{8} + \frac{5 ln ∣ ∣ x^{2} + 4 ∣ ∣}{16} + c$ $((5-ln|x-2|)-(3*tan^-1(x/2)))/8+(5ln|x^2+4|)/16+c$

You can simplify further if you like, however this is an acceptable answer.

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How do you integrate $\int \frac{2 x + 1}{(x - 2) (x^{2} + 4)}$ $int (2x+1) /( (x-2)(x^2+4))$ using partial fractions?

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Explanation:

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How do you integrate $\int \frac{2 x + 1}{(x - 2) (x^{2} + 4)}$ $int (2x+1) /( (x-2)(x^2+4))$ using partial fractions?