Question

Solve the math with explanation.

Don't spamming on it. ​

Answer 1

Answer:

$\frac{x}{1 + ({log \: x)}^{2} } + C \: \: is \: the \: answer$

Step-by-step explanation:

$∫ ( { \frac{ log(x - 1) }{(1 + { log(x) }^{2}) } )}^{2} dx$

Substituting these , it can be written as

$(1) \: \: log(x) = t$

$(2) \: \: x = {e}^{t}$

$(3) \: \: dx = {e}^{t} dt$

Rearranging the equation , we get

$∫ \frac{( {t - 1)}^{2} }{ ({1 + ({t}^{2})) }^{2} } {e}^{t} dt$

solving these ,

$∫ \frac{ {t}^{2} + 1 - 2t }{ {(1 + {t}^{2} )}^{2} } {e}^{t} dt$

$∫ \frac{( {t}^{2} + 1 )- (2t) }{ {(1 + {t}^{2} )}^{2} } {e}^{t} dt$

$∫ \frac{1 + {t}^{2} }{ {(1 + {t}^{2} )}^{2} } - \frac{2t}{ {(1 + {t}^{2} )}^{2} } {e}^{t} dt \: \: \: \: \: \: \: \: \: \: \: \: ...(i)$

Now ,

if f(t) is taken as

$f(t) = ( \frac{1}{1 + {t}^{2} } )$

and , differentiating f(t) with respect to t

$f'(t) = - \frac{2t}{ {(1 + {t}^{2} ) }^{2} }$

Putting both these values in equation ( i )

we get

$∫f(t) + f'(t) {e}^{t} dt$

integrating this

${e}^{t} f(t)$

$\frac{ {e}^{t} }{1 + {t}^{2} } + C$

Putting all values that are taken for substituting

$\frac{x}{1 + ({log \: x)}^{2} } + C$

This is the result.