how to integrate 2x+7/(2x+1)(x+2) with limits 7,0 into ln50

Question

anonymous · Answer

multiply and divide the second term by ((2x+1)-2(x+2))

ganeshie8 · Answer

$\large \int \limits_0^7   \frac{2x+7}{(2x+1)(x+2)} dx$

ganeshie8 · Answer

like that ?

anonymous · Answer

Yup!

ganeshie8 · Answer

familiar with partial fractions ?

anonymous · Answer

You'll need to use partial fractions (C4 Chapter 1) to simplify, then integrate.

anonymous · Answer

Kind of. I got 4/2x+1 -1/x+2

ganeshie8 · Answer

$\large   \frac{2x+7}{(2x+1)(x+2)}  = \frac{A}{2x+1} + \frac{B}{x+2}$

anonymous · Answer

Yup, I got that.

ganeshie8 · Answer

$\large   \frac{2x+7}{(2x+1)(x+2)}  = \frac{4}{2x+1} + \frac{-1}{x+2}$

ganeshie8 · Answer

looks good :)

ganeshie8 · Answer

$\large   \int \limits_0^7    \frac{2x+7}{(2x+1)(x+2)}~dx  =  \int \limits_0^7  \frac{4}{2x+1} + \frac{-1}{x+2}~dx$

ganeshie8 · Answer

and why is that easy to integrate now ?

ganeshie8 · Answer

it is easy cuz we knw :

$\int  \frac{1}{x}~ dx =   \ln x$

ganeshie8 · Answer

you just need to fix the constants okay

ganeshie8 · Answer

$\large   \int \limits_0^7    \frac{2x+7}{(2x+1)(x+2)}~dx  =  \int \limits_0^7  \frac{4}{2x+1} + \frac{-1}{x+2}~dx$


$\large   ~~~~~~~~~~~~~~~~~~=  \int \limits_0^7  \frac{4}{2x+1}~dx + \int \limits_0^7 \frac{-1}{x+2}~dx$

$\large   ~~~~~~~~~~~~~~~~~~=  4\int \limits_0^7  \frac{1}{2x+1}~dx - \int \limits_0^7 \frac{1}{x+2}~dx$

$\large   ~~~~~~~~~~~~~~~~~~=  4  \frac{\ln (2x+1)}{2}\Big|_0^7 - \frac{1}{x+2}\Big|_0^7$

ganeshie8 · Answer

see if that makes sense so far,
next, u just need to take the limits and simplify

anonymous · Answer

Thanks hehe!

ganeshie8 · Answer

u wlc :)
when u take the limits it does simplify to ln50

ganeshie8 · Answer

good luck !

anonymous · Answer

I got ln25 ><

ganeshie8 · Answer

check ur calculation  again

anonymous · Answer

I took (15^2)/2 then /9

ganeshie8 · Answer

$~~~~~~~~~~~~~~~~~~=  4  \frac{\ln (2x+1)}{2}\Big|_0^7 - \ln(x+2)\Big|_0^7$


$~~~~~~~~~~~~~~~~~~=  2   \ln (2x+1)\Big|_0^7 - \ln(x+2)\Big|_0^7$


$~~~~~~~~~~~~~~~~~~=  2  \left[\ln (2*7 + 1) -  \ln(2*0+1)\right]  - \left[\ln(7+2) - \ln(0+2)\right]$


$~~~~~~~~~~~~~~~~~~=  2  \left[\ln (15) -  \ln(1)\right]  - \left[\ln(9) - \ln(2)\right]$


$~~~~~~~~~~~~~~~~~~=  2  \left[\ln (15) \right]  - \left[\ln(9) - \ln(2)\right]$


$~~~~~~~~~~~~~~~~~~=  \ln (15^2) - \ln(9) + \ln(2)$

$~~~~~~~~~~~~~~~~~~=  \ln (\frac{15^2\times 2}{9}) $

$~~~~~~~~~~~~~~~~~~=  \ln (50) $

ganeshie8 · Answer

see if that looks okay...

anonymous · Answer

Omg, thanks!!!! I forgot to separate the two integrals and forgot to sub in 0 ><

ganeshie8 · Answer

np :)