find-x-3-x-2-x-1-dx-

Question Number 35428 by abdo.msup.com last updated on 18/May/18

f i n d \int \frac{x + 3}{\sqrt{x^{2} + x - 1}} d x

Commented by prof Abdo imad last updated on 19/May/18

l e t p u t I = \int \frac{x + 3}{\sqrt{x^{2} + x - 1}} d x

I = \frac{1}{2} \int \frac{2 x + 6}{\sqrt{x^{2} + x - 1}} d x

= \frac{1}{2} \int \frac{2 x + 1}{\sqrt{x^{2} + x - 1}} d x + \frac{5}{2} \int \frac{d x}{\sqrt{x^{2} + x - 1}} b u t

\int \frac{2 x + 1}{2 \sqrt{x^{2} + x - 1}} d x = \sqrt{x^{2} + x - 1} + λ

x^{2} + x - 1 = x^{2} + 2 x \frac{1}{2} + \frac{1}{4} - 1 - \frac{1}{4}

= {(x + \frac{1}{2})}^{2} - \frac{5}{4} a n d c j s n g e m e n t x + \frac{1}{2} = \frac{\sqrt{5}}{2} c h t

g i v e \int \frac{d x}{\sqrt{x^{2} + x - 1}} = \frac{\sqrt{5}}{2} \int \frac{s h t d t}{\sqrt{\frac{5}{4} ({c h}^{2} t - 1)}}

= \frac{\sqrt{5}}{2} . \frac{2}{\sqrt{5}} \int \frac{s h t}{s h t} d t = t + c b u t 2 x + 1 = \sqrt{5} c h t \Rightarrow

c h t = \frac{2 x + 1}{\sqrt{5}} \Rightarrow t = a r g c h (\frac{2 x + 1}{\sqrt{5}})

= l n (\frac{2 x + 1}{\sqrt{5}} + \sqrt{{\frac{2 x + 1}{\sqrt{5}}}^{2} - 1}) \Rightarrow

I = \sqrt{x^{2} + x - 1} + \frac{5}{2} l n {\frac{2 x + 1}{\sqrt{5}} + \sqrt{{(\frac{2 x + 1}{\sqrt{5}})}^{2} - 1}) + λ

Answered by ajfour last updated on 19/May/18

I = \frac{1}{2} \int \frac{2 x + 1}{\sqrt{x^{2} + x - 1}} d x + \frac{5}{2} \int \frac{d x}{\sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{\sqrt{5}}{2})}^{2}}}

= \sqrt{x^{2} + x - 1} + \frac{5}{2} \ln ∣ x + \frac{1}{2} + \sqrt{x^{2} + x - 1} ∣ + c .

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