let-f-x-dt-x-t-t-2-x-2-1-determine-a-explicit-form-of-f-x-2-determine-dt-x-2-t-2-4-and-dt-x-1-t-2-1-

Question Number 59576 by maxmathsup by imad last updated on 12/May/19

l e t f (x) = \int \frac{d t}{(x + t) \sqrt{t^{2} - x^{2}}}

1) d e t e r m i n e a e x p l i c i t f o r m o f f (x)

2) d e t e r m i n e \int \frac{d t}{(x + 2) \sqrt{t^{2} - 4}} a n d \int \frac{d t}{(x + 1) \sqrt{t^{2} - 1}}

Commented by tanmay last updated on 12/May/19

\int \frac{d t}{(t + 2) \sqrt{t^{2} - 4}} a n d \int \frac{d t}{(t + 1) \sqrt{t^{2} - 1}}

\int \frac{d t}{(t + a) \sqrt{t^{2} - a^{2}}}

k = \frac{1}{t + a}

t + a = \frac{1}{k} \to d t = \frac{- d k}{k^{2}}

\int \frac{k}{\sqrt{(\frac{1}{k}) (\frac{1}{k} - 2 a)}} \times \frac{- d k}{k^{2}}

\int \frac{- d k}{k \sqrt{\frac{1 - 2 a k}{k^{2}}}}

p^{2} = 1 - 2 a k \to 2 p d p = - 2 a d k

\int \frac{p d p}{a \times p}

\frac{1}{a} \times p + c

= \frac{\sqrt{1 - 2 a k}}{a} + c

= \frac{\sqrt{1 - \frac{2 a}{t + a}}}{a} + c

n o w p u t a = 2

\frac{\sqrt{1 - \frac{4}{t + 2}}}{2} + c \dots . (a n s f o r Q n o 1)

n o w p u t a = 1

\frac{\sqrt{1 - \frac{2}{t + 1}}}{1} + c \dots . (a n s f o r Q n o 2)

Commented by maxmathsup by imad last updated on 12/May/19

2) t h e Q . i s d e t e r m i n e \int \frac{d t}{(t + 2) \sqrt{t^{2} - 4}} a n d \int \frac{d t}{(t + 1) \sqrt{t^{2} - 1}}

Commented by maxmathsup by imad last updated on 14/May/19

c h a n g e m e n t t = x c h (u) g i v e f (x) = \int \frac{x s h (u)}{(x + x c h (u)) ∣ x ∣ s h (u)} d u

= ξ (x) \int \frac{d u}{x (1 + c h (u))} = \frac{ξ (x)}{x} \int \frac{d u}{1 + \frac{e^{u} + e^{- u}}{2}} = \frac{2 ξ (x)}{x} \int \frac{d u}{2 + e^{u} + e^{- u}}

=_{e^{u} = α} \frac{2 ξ (x)}{x} \int \frac{1}{2 + α + α^{- 1}} \frac{d α}{α} = \frac{2 ξ (x)}{x} \int \frac{d α}{2 α + α^{2} + 1}

= \frac{2 ξ (x)}{x} \int \frac{d α}{{(α + 1)}^{2}} = - \frac{2 ξ (x)}{x} \frac{1}{α + 1} + c = - \frac{2}{x} ξ (x) \frac{1}{e^{u} + 1} b u t

u = a r g c h (\frac{t}{x}) = l n (\frac{t}{x} + \sqrt{\frac{t^{2}}{x^{2}} - 1}) \Rightarrow e^{u} = \frac{t}{x} + \sqrt{\frac{t^{2}}{x^{2}} - 1} \Rightarrow

f (x) = - \frac{2}{x} ξ (x) \frac{1}{\frac{t}{x} + \frac{\sqrt{t^{2} - x^{2}}}{∣ x ∣}} + c = - \frac{2 ξ (x)}{t + ξ (x) \sqrt{t^{2} - x^{2}}} + c w i t h

ξ (x) = 1 i f x > 0 a n d ξ (x) = - 1 i f x < 0 s o

f (x) = \frac{- 2}{t + \sqrt{t^{2} - x^{2}}} + c i f x > 0

f (x) = \frac{2}{t - \sqrt{t^{2} - x^{2}}} + c i f x < 0

Commented by maxmathsup by imad last updated on 14/May/19

2) \int \frac{d t}{(t + 2) \sqrt{t^{2} - 4}} = f (2) = \frac{- 2}{t + \sqrt{t^{2} - 4}} + c

\int \frac{d t}{(t + 1) \sqrt{t^{2} - 1}} = f (1) = \frac{- 2}{t + \sqrt{t^{2} - 1}} + c