How do you integrate $\frac{2 x + 3}{x^{2} - 9}$ $(2x+3)/(x^2-9)$ using partial fractions?

Question

How do you integrate $\frac{2 x + 3}{x^{2} - 9}$ $(2x+3)/(x^2-9)$ using partial fractions?

1 Answer

Impact of this question

7941 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-17T00:16:25

$\int \frac{2 x + 3}{x^{2} - 9} d x = \frac{ln | x + 3 | + 3 ln | x - 3 |}{2} + C$ $int(2x+3)/(x^2-9)dx=(lnabs(x+3)+3lnabs(x-3))/2+C$

Explanation:

Since the numerator is in a lower degree than the denominator, we can proceed to the next step.

$\int \frac{2 x + 3}{x^{2} - 9} d x$ $int(2x+3)/(x^2-9)dx$ Factor

$\int \frac{2 x + 3}{(x + 3) (x - 3)} d x$ $int(2x+3)/((x+3)(x-3))dx$

We can rewrite the rational function in the form:

$\frac{A}{x + 3} + \frac{B}{x - 3}$ $A/(x+3)+B/(x-3)$

$\Rightarrow \frac{A}{x + 3} \cdot \frac{x - 3}{x - 3} + \frac{B}{x - 3} \cdot \frac{x + 3}{x + 3}$ $=>A/(x+3)* (x-3)/(x-3)+B/(x-3)*(x+3)/(x+3)$

$\Rightarrow \frac{A (x - 3)}{(x + 3) (x - 3)} + \frac{B (x + 3)}{(x - 3) (x + 3)}$ $=>(A(x-3))/((x+3)(x-3))+(B(x+3))/((x-3)(x+3))$

$\Rightarrow \frac{A (x - 3) + B (x + 3)}{(x - 3) (x + 3)}$ $=>(A(x-3)+B(x+3))/((x-3)(x+3))$

$\Rightarrow \frac{A x - 3 A + B x + 3 B}{(x - 3) (x + 3)}$ $=>(Ax-3A+Bx+3B)/((x-3)(x+3))$

$\Rightarrow \frac{A x + B x + 3 B - 3 A}{(x - 3) (x + 3)}$ $=>(Ax+Bx+3B-3A)/((x-3)(x+3))$

$\Rightarrow \frac{x (A + B) + 3 (B - A)}{(x - 3) (x + 3)} = \frac{2 x + 3}{(x + 3) (x - 3)}$ $=>(x(A+B)+3(B-A))/((x-3)(x+3))=(2x+3)/((x+3)(x-3))$

$\Rightarrow x (A + B) + 3 (B - A) = 2 x + 3$ $=>x(A+B)+3(B-A)=2x+3$

We let:

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

$3 (B - A) = 3$ $3(B-A)=3$

$\Rightarrow A + B = 2$ $=>A+B=2$

$\Rightarrow - A + B = 1$ $=>-A+B=1$ Add the equations together.

$\Rightarrow 2 B = 3$ $=>2B=3$

$\Rightarrow B = \frac{3}{2}$ $=>B=3/2$

$\Rightarrow A = \frac{1}{2}$ $=>A=1/2$

We substitute this in the original form to get:

$\int \frac{1}{2} \cdot \frac{1}{x + 3} + \frac{3}{2} \cdot \frac{1}{x - 3} d x$ $int1/2*1/(x+3)+3/2*1/(x-3)dx$

$\Rightarrow \int \frac{1}{2} \cdot \frac{1}{x + 3} d x + \int \frac{3}{2} \cdot \frac{1}{x - 3} d x$ $=>int1/2*1/(x+3)dx+int3/2*1/(x-3)dx$

$\Rightarrow \frac{1}{2} \int \frac{1}{x + 3} d x + \frac{3}{2} \int \frac{1}{x - 3} d x$ $=>1/2int1/(x+3)dx+3/2int1/(x-3)dx$

Remember that:

$\int \frac{1}{x + n} d x = ln | x + n |$ $int1/(x+n)dx=lnabs(x+n)$

$\Rightarrow \frac{1}{2} \cdot ln | x + 3 | + \frac{3}{2} ln | x - 3 |$ $=>1/2*lnabs(x+3)+3/2lnabs(x-3)$

$\Rightarrow \frac{ln | x + 3 |}{2} + \frac{3 ln | x - 3 |}{2}$ $=>lnabs(x+3)/2+(3lnabs(x-3))/2$

$\Rightarrow \frac{ln | x + 3 | + 3 ln | x - 3 |}{2}$ $=>(lnabs(x+3)+3lnabs(x-3))/2$ Do you $C$ $C$ why this is incomplete?

$\Rightarrow \frac{ln | x + 3 | + 3 ln | x - 3 |}{2} + C$ $=>(lnabs(x+3)+3lnabs(x-3))/2+C$

That is the answer!

Science

Math

Humanities

... and beyond

How do you integrate $\frac{2 x + 3}{x^{2} - 9}$ $(2x+3)/(x^2-9)$ using partial fractions?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate 2x+3x2−9(2x+3)/(x^2-9) using partial fractions?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\frac{2 x + 3}{x^{2} - 9}$ $(2x+3)/(x^2-9)$ using partial fractions?