find-dx-x-n-x-1-m-m-and-n-integr-

Question Number 95586 by turbo msup by abdo last updated on 26/May/20

f i n d \int \frac{d x}{x^{n} {(x + 1)}^{m}}

m a n d n i n t e g r

Commented by Tony Lin last updated on 26/May/20

l e t u = \frac{x}{x + 1}, d x = {(x + 1)}^{2} d u

x = \frac{u}{1 - u}

\int \frac{1}{{(\frac{u}{1 - u})}^{n} {(\frac{1}{1 - u})}^{m - 2}} d u

= \int \frac{{(1 - \frac{x}{x + 1})}^{n + m - 2}}{{(\frac{x}{x + 1})}^{n}} d (\frac{x}{x + 1})

= C_{n - 1}^{n + m - 2} {(- 1)}^{n - 1} l n ∣ \frac{x}{x + 1} ∣

+ \underset{k = 0}{\sum^{n - 2}} C_{k}^{n + m - 2} {(- 1)}^{k + 1} \frac{1}{(n + k - 1) {(\frac{x}{x + 1})}^{u + k - 1}}

+ \underset{k = 0}{\sum^{m - 2}} C_{n + k}^{n + m - 2} {(- 1)}^{n + k} \frac{{(\frac{x}{x + 1})}^{k + 1}}{k + 1}

+ c

Answered by mathmax by abdo last updated on 26/May/20

I = \int \frac{dx}{x^{n} {(x + 1)}^{m}} \Rightarrow I = \int \frac{dx}{{(\frac{x}{x + 1})}^{n} {(x + 1)}^{m + n}}

we do the changement \frac{x}{x + 1} = t \Rightarrow x = tx + t \Rightarrow (1 - t) x = t

\Rightarrow x = \frac{t}{1 - t} \Rightarrow \frac{dx}{dt} = \frac{1 - t - t (- 1)}{{(1 - t)}^{2}} = \frac{1}{{(1 - t)}^{2}} and x + 1 = \frac{t}{1 - t} + 1 = \frac{1}{1 - t}

I = \int \frac{dt}{{(1 - t)}^{2} t^{n} {(\frac{1}{1 - t})}^{m + n}} = \int \frac{{(1 - t)}^{m + n}}{{(1 - t)}^{2} t^{n}} dt

= \int \frac{{(1 - t)}^{m + n - 2}}{t^{n}} dt = {(- 1)}^{m + n - 2} \int \frac{{(t - 1)}^{m + n - 2}}{t^{n}} dt

= {(- 1)}^{m + n - 2} \int \frac{\sum_{k = 0}^{m + n - 2} C_{m + n - 2}^{k} t^{k}}{t^{n}} dt

= {(- 1)}^{m + n - 2} \sum_{k = 0}^{m + n - 2} C_{m + n - 2}^{k} \int t^{k - n} dt

= {(- 1)}^{m + n - 2} \sum_{k = 0 and k \neq n - 1}^{m + n - 2} C_{m + n - 2}^{k} \frac{1}{k - n + 1} t^{k - n + 1}

+ {(- 1)}^{m + n - 2} C_{m + n - 2}^{n - 1} \ln ∣ t ∣ + C

\Rightarrow I = {(- 1)}^{m + n - 2} \sum_{k = 0 and k \neq n - 1}^{m + n - 2} C_{m + n - 2}^{k} \frac{1}{k - n + 1} {(\frac{x}{x + 1})}^{k - n + 1}

+ {(- 1)}^{m + n - 2} C_{m + n - 2}^{n - 1} \ln ∣ \frac{x}{x + 1} ∣ + C