Question

Partial Fractions mistake this time...posting in a minute

Answer 1

\[\large \int\limits_{}^{}\frac{5x - 4}{(2x + 1)(x - 3)}\]
>Broken into partial fractions
\[\large 5x - 4 = \frac{A}{(2x + 1)} + \frac{B}{(x - 3)}\]
\[\large 5x - 4 = A(x - 3) + B(2x + 1)\]

(x - 3) = 0 when x = 3

When x = 3 we have

\[\large 5(3) - 4 = A((3)-3) + B(2(3) + 1)\]
\[\large 11 = 7B\]
\[\large B = \frac{11}{7}\]

Now for the other side

(2x + 1) = 0 when x = -1/2

When x = -1/2 we have

\[\large 5(-\frac{1}{2}) - 4 = A((-\frac{1}{2})-3) + B(2(-\frac{1}{2}) + 1)\]
\[\large (-\frac{5}{2}) - 4 = A(-\frac{7}{2})\]
\[\large -\frac{13}{2} = -\frac{7}{2}A\]
\[\large A = \frac{26}{14}\]
\[\large A = \frac{13}{7}\]

So altogether I have

\[\large \int\limits_{}^{}\frac{\frac{13}{7}}{2x + 1}dx + \int\limits_{}^{} \frac{\frac{11}{7}}{x - 3}dx\]

>After obvious u-subs and use of ln...

\[\large \frac{13}{7}\ln(2x + 1) + \frac{11}{7}\ln(x - 3)\]

NOW my textbook says THIS answer is supposed to be

\[\large \frac{13}{14}\ln(2x + 1) + \frac{11}{7}\ln(x - 3)\]

Once again I cannot find where I went off the track...and I know it's going to be a simple mistake again...

Answer 2

I just can't find it @ganeshie8 lol

Answer 3

partial fractions are correct,.

mistake is at u-sub in first integral

Answer 4

2x+1 = u

2 dx = du

Answer 5

AH!!! yes see such a simple mistake! Thank you @ganeshie8 !

Answer 6

np :)