# Thread: partial fractions and integration

1. ## partial fractions and integration

Express (4x-13)/[(x+2)(2x-3)] in partial fractions. Hence or otherwise, evaluate \int_2^3 [(4x-13)/{(x+2)(2x-3)}\ dx

2. Originally Posted by maybeline9216
Express (4x-13)/[(x+2)(2x-3)] in partial fractions. Hence or otherwise, evaluate int_2^3 [(4x-13)/{(x+2)(2x-3)} dx
$\dfrac{4x-13}{(x+2)(2x-3)} = \dfrac{A}{x+2} + \dfrac{B}{2x-3} = \dfrac{(2A+B)x - 3A + 2B}{(x+2)(2x-3)}$

That means you have to solve the system of simultaneous equations:

$\left|\begin{array}{l}2A+B=4 \\ -3A+2B = -13\end{array}\right.$ ......... which will yield A = 3 and B = -2

3. Originally Posted by earboth
$\dfrac{4x-13}{(x+2)(2x-3)} = \dfrac{A}{x+2} + \dfrac{B}{2x-3} = \dfrac{(2A+B)x - 3A + 2B}{(x+2)(2x-3)}$

That means you have to solve the system of simultaneous equations:

$\left|\begin{array}{l}2A+B=4 \\ -3A+2B = -13\end{array}\right.$ ......... which will yield A = 3 and B = -2
Thanks...but i know how to do the 1st part of the question..its just that my answer is different from my paper's answer for the 2nd part of the question(integrating part)

4. Originally Posted by maybeline9216
Express (4x-13)/[(x+2)(2x-3)] in partial fractions. Hence or otherwise, evaluate \int_2^3 [(4x-13)/{(x+2)(2x-3)}\ dx
$\int_2^3\dfrac{4x-13}{(x+2)(2x-3}dx= \int_2^3\dfrac{3}{x+2}dx - \int_2^3\dfrac{2}{2x-3}dx$ = $\left. 3\ln(x+2)\right|_2^3 - \left. \ln(2x-3)\right|_2^3 = \ln\left(\dfrac{125}{192} \right)$

5. Originally Posted by earboth
$\int_2^3\dfrac{4x-13}{(x+2)(2x-3}dx= \int_2^3\dfrac{3}{x+2}dx - \int_2^3\dfrac{2}{2x-3}dx$ = $\left. 3\ln(x+2)\right|_2^3 - \left. \ln(2x-3)\right|_2^3 = \ln\left(\dfrac{125}{192} \right)$
My workings same as you but...anyway u missed out the 2

Is that = 3(ln5-ln4)-2(ln3-ln1) ?

6. Originally Posted by maybeline9216
My workings same as you but...anyway u missed out the 2

Is that = 3(ln5-ln4)-2(ln3-ln1) ?
$\int\dfrac{2}{2x-3}dx$

Let $u = 2x-3~\implies~\frac{du}{dx}=2~\implies~dx=\frac{du}2$

Then you have:

$\int \dfrac{2}{u}\frac{du}2 = \int \frac1u du = \ln(u) = \ln(2x-3)$

7. Originally Posted by earboth
$\int\dfrac{2}{2x-3}dx$

Let $u = 2x-3~\implies~\frac{du}{dx}=2~\implies~dx=\frac{du}2$

Then you have:

$\int \dfrac{2}{u}\frac{du}2 = \int \frac1u du = \ln(u) = \ln(2x-3)$
In that case, why is it that this doesn't apply to
$\int_2^3\dfrac{3}{x+2}dx$ and why is it that i cannot just take out the 2??

8. Originally Posted by maybeline9216
In that case, (A) why is it that this doesn't apply to
$\int_2^3\dfrac{3}{x+2}dx$ and (B) why is it that i cannot just take out the 2??

to (A): The 3 in the numerator of the quotient has nothing to do with the variable x. It is a constant factor which can be taken in front of the integral.

(B) The 2 in the numerator is the derivate of the denominator and is used to do the integration by substitution.

Use the search function of the forum and look for examples of integration by substitution. For instance Krizalid has posted a lot of examples.