Question

$\displaystyle \bold \red{\int^{4}_{2} \frac{ \log( {x}^{2}) }{ \log( {x}^{2} ) + \log(36 - 12x + {x}^{2} )} \: dx}$



❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​

Answer 1

${\displaystyle \bold \red{I=\int^{4}_{2} \frac{ \log{x}^{2} }{ \log{x}^{2} + \log(36 - 12x + {x}^{2} )} \: dx}}$

${\displaystyle \bold \red{=\int^{4}_{2} \frac{ \log{x}^{2} }{ \log{x}^{2} + \log(6 - x ){}^{2}} \: dx}}$

${\displaystyle \bold \red{=\int^{4}_{2} \frac{ \log{x}^{} }{ \log{x}^{} + \log(6 - x ){}^{}} \: dx - - - - (1)}}$

${\displaystyle \bold \red{ \implies I=\int^{4}_{2} \frac{ \log{(2 + 4 - x)}^{} }{ \log{(2 + 4 - x)}^{} + \log(x)} \: dx}}$

By using property ${ \displaystyle \bold{ \int^{b}_{a}f(x)dx }}$

$\displaystyle{ \bold \red{ = \int^{b}_{a}f(a + b - x)dx,\: we \: get}}$

${\displaystyle \bold \red{ \implies I=\int^{4}_{2} \frac{ \log{(6 - x)}^{} }{ \log{(6 - x)}^{} + \log x} \: - - - - (2)}}$

Eqs. (1)+(2) gives,

${\displaystyle \bold \red{ 2I =\int^{4}_{2} 1 \: dx = 2}}$

${\displaystyle \bold \red{ \implies I =1}}$