Question

Let:-
$\int \frac{dx}{ {x}^{2008} + x} = \frac{1}{p} ln( \frac{{x}^{q}}{1 + {x}^{r} } ) + c$
Find p+q+r ​

Answer 1

The appearance can be converted as follows:-

$\LARGE \int \frac{dx}{x( {x}^{n} + 1) } = \frac{1}{n} \int \frac{n. {x}^{n - 1} dx}{{x}^{n - 1 }x( {x}^{n} + 1) }$

Now call $x^n=t$

$nx^{n-1}dx=dt$

$: \rightarrow \: \: \LARGE \: \frac{1}{n} \int \frac{dt}{ {x}^{n} ( {x}^{n} + 1) }$

$: \rightarrow \: \: \LARGE \: \frac{1}{n} \int \frac{dt}{ t ( t + 1) }$

$: \rightarrow \: \: \LARGE \: \frac{1}{n} (\int \frac{dt}{ t } - \int \frac{dt}{( t + 1)})$

$: \rightarrow \: \: \: \LARGE \mathtt{\frac{1}{n}( ln |t| - (ln( |t + 1| + c ) }$

We substituted $x^n=t$ now putting value of $t \: as \: x^n$.

$: \rightarrow \: \: \: \LARGE \mathtt{\frac{1}{n}( ln | {x}^{n} | - (ln( | {x}^{n}+ 1| + c ) }$

$: \rightarrow \: \: \: \LARGE \frac{1}{n}( ln | \frac{{x}^{n}}{{x}^{n} + 1} | + c )$

And we know that ${x}^{n} = x^{2007}$

Hence n is 2007

By comparing with the equation given in equation we get p,q,r=2007

Hence p+q+r=$\boxed{6021}$