find-the-value-of-0-1-dx-x-2-2x-5-

Question Number 26396 by abdo imad last updated on 25/Dec/17

f i n d t h e v a l u e o f \int_{0}^{1} \frac{d x}{x^{2} + 2 x + 5} .

Commented by abdo imad last updated on 26/Dec/17

l e t p u t I = \int_{0}^{1} \frac{d x}{x^{2} + 2 x + 5} d u e t o x^{2} + 2 x + 5 = {(x + 1)}^{2} + 4 w e u s e

t h e c h a n g e m e n t x + 1 = 2 t \Rightarrow I = \int_{\frac{1}{2}}^{1} \frac{2 d t}{4 t^{2} + 4} = \frac{1}{2} \int_{\frac{1}{2}}^{1} \frac{d t}{1 + t^{2}}

I = \frac{1}{2} {[a r c t a n t]}_{\frac{1}{2}}^{1} = \frac{1}{2} (a r c t a n (1) - a r c t a n (\frac{1}{2}))

\Rightarrow I = \frac{π}{8} - \frac{1}{2} a r c t a n (\frac{1}{2})

Answered by jota@ last updated on 25/Dec/17

\int_{0}^{1} \frac{d (x + 1)}{{(x + 1)}^{2} + 2^{2}} = \frac{1}{2} \tan^{- 1} (\frac{x + 1}{2}) ∣_{0}^{1}

= \frac{1}{2} [\tan^{- 1} 1 - \tan^{- 1} (1 / 2)] .

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