find-the-value-of-0-dx-x-1-x-2-x-3-

Question Number 26759 by abdo imad last updated on 29/Dec/17

f i n d t h e v a l u e o f \int_{0}^{\propto} \frac{d x}{(x + 1) (x + 2) (x + 3)} .

Commented by abdo imad last updated on 30/Dec/17

w e u s e t h e c h a n g e m e n t x + 2 = t

I = \int_{0}^{\propto} \frac{d x}{(x + 1) (x + 2) (x + 3)} = \int_{2}^{\propto} \frac{d t}{(t - 1) t (t + 1)} a n d a f t e r d e c o m p o s i t i o n

I = \int_{2}^{\propto} (\frac{1}{2 (t - 1)} - \frac{1}{t} + \frac{1}{2 (t + 1)}) d t

= \frac{1}{2} \int_{2}^{\propto} (\frac{1}{t - 1} + \frac{1}{t + 1}) d t - \int_{2}^{\propto} \frac{d t}{t}

= \frac{1}{2} \int_{2}^{\propto} \frac{2 t}{t^{2} - 1} d t - \int_{2}^{\propto} \frac{d t}{t} = {[\frac{1}{2} l n / t^{2} - 1 / - l n / t /]}_{2}^{\propto}

= [l n {(\frac{\sqrt{/ t^{2} - 1 /}}{/ t /}]}_{2}^{\propto} = {[l n (\frac{\sqrt{t^{2} - 1}}{t})]}_{2}^{\propto} = - l n (\frac{\sqrt{3}}{2})

I = l n (2) - \frac{1}{2} l n (3) .

Answered by ajfour last updated on 29/Dec/17

\frac{1}{(x + 1) (x + 2) (x + 3)} = \frac{1}{t (t^{2} - 1)}

i f x + 2 = t

l e t t = \sec^{2} θ \Rightarrow d t = 2 \sec^{2} θ \tan θ

\int_{2}^{α + 2} \frac{d t}{t (t^{2} - 1)} = 2 \int_{θ_{1}}^{θ_{2}} \cot θ d θ

= 2 \ln ∣ \frac{\sin θ_{2}}{\sin θ_{1}} ∣= \ln (\frac{1 - \frac{1}{α + 2}}{1 - \frac{1}{2}})

= \ln (2 - \frac{2}{α + 2}) .

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