find-x-2-x-2-4x-3-dx-

Question Number 48719 by Abdo msup. last updated on 27/Nov/18

f i n d \int \frac{x - 2}{\sqrt{x^{2} + 4 x - 3}} d x

Commented by Abdo msup. last updated on 27/Nov/18

I = \int \frac{x - 2}{\sqrt{x^{2} + 4 x + 4 - 7}} = \int \frac{x - 2}{\sqrt{{(x + 2)}^{2} - 7}} d x

=_{x + 2 = \sqrt{7} c h (t)} \int \frac{\sqrt{7} c h (t) - 4}{\sqrt{7} s h (t)} \sqrt{7} s h (t) d t

= \sqrt{7} \int c h (t) - 4 t + c = \sqrt{7} s h (t) - 4 t + c b u t

s h (t) = \frac{e^{t} - e^{- t}}{2} a n d t = a r g c h (\frac{x + 2}{\sqrt{7}})

= l n (x + \sqrt{{(\frac{x + 2}{\sqrt{7}})}^{2} - 1}) \Rightarrow s h (t) = \frac{x + \sqrt{{(\frac{x + 2}{\sqrt{7}})}^{2} - 1} - \frac{1}{x + \sqrt{(\frac{x + 2}{\sqrt{7}}) - 1}}}{2}

I = \frac{\sqrt{7}}{2} {(x + \sqrt{{(\frac{x + 2}{\sqrt{7}})}^{2} - 1} - {(x + \sqrt{{(\frac{x + 2}{\sqrt{7}})}^{2} - 1})}^{- 1}} - 4 t + c .

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18

t^{2} = x^{2} + 4 x - 3

2 t \frac{d t}{d x} = 2 x + 4

t d t = (x + 2) d x

\int \frac{x + 2 - 4}{\sqrt{x^{2} + 4 x - 3}} d x

\int \frac{t d t}{t} - 4 \int \frac{d x}{\sqrt{{(x + 2)}^{2} - {(\sqrt{7})}^{2}}}

t - 4 l n {(x + 2) + \sqrt{{(x + 2)}^{2} - {(\sqrt{7})}^{2}}} + c

\sqrt{(x^{2} + 4 x - 3)} - 4 l n {(x + 2) + \sqrt{x^{2} + 4 x - 3}} + c