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Question Number 37018 by behi83417@gmail.com last updated on 08/Jun/18

Commented by math khazana by abdo last updated on 08/Jun/18

$1) l e t I = \int \frac{\sqrt{1 + x^{2}}}{1 + x} d x c h a n g e m e n t x = s h t g i v e$ $I = \int \frac{c h (t)}{1 + s h (t)} c h (t) d t = \int \frac{{c h}^{2} (t) d t}{1 + s h (t)}$ $= \frac{1}{2} \int \frac{1 + c h (2 t)}{1 + s h (t)} d t = \frac{1}{2} \int \frac{1 + \frac{e^{2 t} + e^{- 2 t}}{2}}{1 + \frac{e^{t} - e^{- t}}{2}} d t$ $2 I = \int \frac{2 + e^{2 t} + e^{- 2 t}}{2 + e^{t} - e^{- t}} d t$ $=_{e^{t} = u} \int \frac{2 + u^{2} + \frac{1}{u^{2}}}{2 + u - \frac{1}{u}} d u = \int \frac{2 u^{2} + u^{4} + 1}{2 u + u^{2} - 1} \frac{1}{u} d u$ $= \int \frac{u^{4} + 2 u^{2} + 1}{u^{3} + 2 u^{2} - u} d u$ $= \int \frac{u (u^{3} + 2 u^{2} - u) - 2 u^{3} + u^{2} + 2 u^{2} + 1}{u^{3} + 2 u^{2} - u} d u$ $= \frac{u^{2}}{2} + \int \frac{- 2 u^{3} + 3 u^{2} + 1}{u^{3} + 2 u^{2} - u} d u$ $= \frac{u^{2}}{2} + \int \frac{- 2 (u^{3} + 2 u^{2} - u) + 4 u^{2} - 2 u + 3 u^{2} + 1}{u^{3} + 2 u^{2} - u} d u$ $= \frac{u^{2}}{2} - 2 u + \int \frac{7 u^{2} - 2 u + 1}{u (u^{2} + 2 u - 1)} d u l e t ? d e c o m p o s e$ $F (u) = \frac{7 u^{2} - 2 u + 1}{u (u^{2} + 2 u - 1)} r o o t s o f u^{2} + 2 u - 1$ $Δ^{^{'}} = 2 \Rightarrow u_{1} = - 1 + \sqrt{2} a n d u_{2} = - 1 - \sqrt{2}$ $F (u) = \frac{7 u^{2} - 2 u + 1}{u (u - u_{1}) (u - u_{2})} = \frac{a}{u} + \frac{b}{u - u_{1}} + \frac{c}{u - u_{2}}$ $i t s s i m p l e t o f i n d a, b a n d c s o$ $\int F (u) d u = a l n ∣ u ∣ + b l n ∣ u - u_{1} ∣ + c l n ∣ u - u_{2} ∣ + C$ $= a t + b l n ∣ e^{t} - u_{1} ∣ + c l n ∣ e^{t} - u_{2} ∣ + C$ $= a l n (x + \sqrt{1 + x^{2}}) + b l n ∣ x + \sqrt{1 + x^{2}} - u_{1} ∣$ $+ c l n ∣ x + \sqrt{1 + x^{2}} - u_{2} ∣ + C = 2 I \Rightarrow$ $I = \frac{a}{2} l n (x + \sqrt{1 + x^{2}}) + \frac{b}{2} l n ∣ x + \sqrt{1 + x^{2}} - u_{1} ∣$ $+ \frac{c}{2} l n ∣ x + \sqrt{1 + x^{2}} - u_{2} ∣ + C .$
Commented by behi83417@gmail.com last updated on 08/Jun/18

$\int \frac{\sqrt{1 + x^{2}}}{1 + x} d x =$ $\sqrt{2} l n (\frac{\sqrt{1 + x^{2}} - x - \sqrt{2} - 1}{\sqrt{1 + x^{2}} - x + \sqrt{2} - 1}) + l n (\sqrt{1 + x^{2}} - x) + \sqrt{1 + x^{2}} + C$
Commented by MJS last updated on 08/Jun/18

$sorry but this is wrong$ $\int \frac{\sqrt{1 + x^{2}}}{1 + x} d x =$ $[t = \arctan x \to d x = \sec^{2} t d t]$ $= \int \frac{\sec^{2} t \sqrt{1 + \tan^{2} t}}{1 + \tan t} d t = \int \frac{\sec^{3} t}{1 + \tan t} =$ $[Weierstrass : u = \tan \frac{t}{2} \to d t = \frac{2 d u}{1 + u^{2}}]$ $= - 2 \int \frac{{(u^{2} + 1)}^{2}}{{(u - 1)}^{2} {(u + 1)}^{2} (u^{2} - 2 u - 1)} d u =$ $= - 4 \int \frac{d u}{u^{2} - 2 u - 1} +$ $+ \int \frac{d u}{{(u - 1)}^{2}} - \int \frac{d u}{{(u + 1)}^{2}} +$ $+ \int \frac{d u}{u - 1} - \int \frac{d u}{u + 1} =$ $= - \sqrt{2} \int \frac{d u}{u - \sqrt{2} - 1} + \sqrt{2} \int \frac{d u}{u + \sqrt{2} - 1} -$ $- \frac{1}{u - 1} + \frac{1}{u + 1} +$ $+ \ln (u - 1) - \ln (u + 1) =$ $= - \sqrt{2} \ln (u - \sqrt{2} - 1) + \sqrt{2} \ln (u + \sqrt{2} - 1) +$ $+ \frac{2}{1 - u^{2}} + \ln \frac{u - 1}{u + 1} =$ $= \sqrt{2} \ln \frac{u + \sqrt{2} - 1}{u - \sqrt{2} - 1} + \ln \frac{u - 1}{u + 1} + \frac{2}{1 - u^{2}} =$ $= \sqrt{2} \ln \frac{\tan \frac{t}{2} + \sqrt{2} - 1}{\tan \frac{t}{2} - \sqrt{2} - 1} + \ln \frac{\tan \frac{t}{2} - 1}{\tan \frac{t}{2} + 1} + \frac{2}{1 - \tan^{2} \frac{t}{2}} =$ $[\tan \frac{\arctan x}{2} = \frac{\sqrt{x^{2} + 1} - 1}{x}]$ $= \sqrt{2} \ln \frac{\sqrt{x^{2} + 1} - (1 - \sqrt{2}) x - 1}{\sqrt{x^{2} + 1} - (1 + \sqrt{2}) x - 1} + \ln \frac{\sqrt{x^{2} + 1} - x - 1}{\sqrt{x^{2} + 1} + x - 1} + \frac{x^{2}}{\sqrt{x^{2} + 1} - 1} =$ $= \sqrt{2} \ln \frac{\sqrt{x^{2} + 1} - (1 - \sqrt{2}) x - 1}{\sqrt{x^{2} + 1} - (1 + \sqrt{2}) x - 1} + \ln (x - \sqrt{x^{2} + 1}) + \sqrt{x^{2} + 1} + 1 =$ $= \sqrt{2} \ln \frac{(1 - \sqrt{2}) (x - 1) - (2 - \sqrt{2}) \sqrt{x^{2} + 1}}{x + 1} + \ln (x - \sqrt{x^{2} + 1}) + \sqrt{x^{2} + 1} + C$
	Answered by MJS last updated on 08/Jun/18

	$2)$ $\int \frac{\sqrt{x}}{x^{3} + x^{2} + x + 1} d x =$ $[t = \sqrt{x} \to d x = 2 \sqrt{x} d t]$ $= 2 \int \frac{t^{2}}{t^{6} + t^{4} + t^{2} + 1} d t = 2 \int \frac{t^{2}}{(t^{4} + 1) (t^{2} + 1)} d t =$ $= \int \frac{t^{2} + 1}{t^{4} + 1} d t - \int \frac{d t}{t^{2} + 1} = \int \frac{t^{2} + 1}{t^{4} + 1} d t - \arctan t =$ $\int \frac{t^{2} + 1}{t^{4} + 1} d t = \int \frac{t^{2} + 1}{(t^{2} - \sqrt{2} t + 1) (t^{2} + \sqrt{2} t + 1)} d t =$ $= \frac{1}{2} \int \frac{d t}{t^{2} - \sqrt{2} t + 1} + \frac{1}{2} \int \frac{d t}{t^{2} + \sqrt{2} t + 1} =$ $= \frac{1}{2} \int \frac{d t}{{(t - \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}} + \frac{1}{2} \int \frac{d t}{{(t + \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}} =$ $[u = \sqrt{2} t \pm 1 \to d t = \frac{\sqrt{2}}{2} d u]$ $= \frac{\sqrt{2}}{2} \int \frac{{d u}_{-}}{u_{-}^{2} + 1} + \frac{\sqrt{2}}{2} \int \frac{{d u}_{+}}{u_{+}^{2} + 1} =$ $= \frac{\sqrt{2}}{2} \arctan u_{-} + \frac{\sqrt{2}}{2} \arctan u_{+} =$ $= \frac{\sqrt{2}}{2} \arctan (\sqrt{2} t - 1) + \frac{\sqrt{2}}{2} \arctan (\sqrt{2} t + 1)$ $= \frac{\sqrt{2}}{2} \arctan (\sqrt{2} t - 1) + \frac{\sqrt{2}}{2} \arctan (\sqrt{2} t + 1) - \arctan t =$ $= \frac{\sqrt{2}}{2} (\arctan (\sqrt{2 x} - 1) + \arctan (\sqrt{2 x} + 1)) - \arctan \sqrt{x} + C$
	Answered by MJS last updated on 08/Jun/18

	$1) I messed around a little, found this :$ $\int \frac{\sqrt{1 + x^{2}}}{1 + x} d x =$ $= \sqrt{2} arcsinh (- \frac{1 - x}{∣ 1 + x ∣}) - arcsinh x + \sqrt{1 + x^{2}} + C$ $let me explain :$ $\int \frac{\sqrt{1 + x^{2}}}{1 + x} d x = \int \frac{1 + x^{2}}{(1 + x) \sqrt{1 + x^{2}}} d x = \int \frac{2 + x^{2} - 1}{(1 + x) \sqrt{1 + x^{2}}} d x =$ $= 2 \int \frac{d x}{(1 + x) \sqrt{1 + x^{2}}} - \int \frac{1 - x^{2}}{(1 + x) \sqrt{1 + x^{2}}} d x =$ $= 2 \int \frac{d x}{(1 + x) \sqrt{1 + x^{2}}} - \int \frac{(1 - x) (1 + x)}{(1 + x) \sqrt{1 + x^{2}}} d x =$ $= 2 \int \frac{d x}{(1 + x) \sqrt{1 + x^{2}}} - \int \frac{1 - x}{\sqrt{1 + x^{2}}} d x =$ $= 2 \int \frac{d x}{(1 + x) \sqrt{1 + x^{2}}} - \int \frac{d x}{\sqrt{1 + x^{2}}} + \int \frac{x}{\sqrt{1 + x^{2}}} d x =$ $[the 2^{nd} and 3^{rd} ones are standard$ $integrals]$ $= 2 \int \frac{d x}{(1 + x) \sqrt{1 + x^{2}}} - arcsinh x + \sqrt{1 + x^{2}}$ $now consider this :$ $\frac{d}{d x} [arcsinh x] = \frac{1}{\sqrt{1 + x^{2}}}$ $\frac{d}{d x} [arcsinh f (x)] = \frac{f^{'} (x)}{\sqrt{1 + f^{2} (x)}}$ $let f (x) = \frac{g (x)}{h (x)} = \frac{u}{v} \Rightarrow$ $\Rightarrow \frac{f^{'} (x)}{\sqrt{1 + f^{2} (x)}} = \frac{\frac{u^{'} v - {u v}^{'}}{v^{2}}}{\sqrt{1 + {(u / v)}^{2}}} = \frac{u^{'} v - {u v}^{'}}{∣ v ∣ \sqrt{u^{2} + v^{2}}}$ $let v = 1 + x \Rightarrow$ $\Rightarrow \frac{u^{'} v - {u v}^{'}}{∣ v ∣ \sqrt{u^{2} + v^{2}}} = \frac{u^{'} (1 + x) - u}{∣ 1 + x ∣ \sqrt{u^{2} + 1 + 2 x + x^{2}}}$ ${we}^{'} d like the following equations to be true :$ $(A) u^{'} (1 + x) - u = 2$ $(B) \sqrt{u^{2} + 1 + 2 x + x^{2}} = \sqrt{1 + x^{2}}$ $(A) works with u = x - 1 but (B) \sqrt{u^{2} + 1 + 2 x + x^{2}} =$ $= \sqrt{{(x - 1)}^{2} + 1 + 2 x + x^{2}} = \sqrt{2 (1 + x^{2})}$ $we now have :$ $\frac{d}{d x} [arcsinh (- \frac{1 - x}{1 + x})] = \frac{\sqrt{2}}{∣ 1 + x ∣ \sqrt{1 + x^{2}}} \Rightarrow$ $\Rightarrow 2 \int \frac{d x}{∣ 1 + x ∣ \sqrt{1 + x^{2}}} = \sqrt{2} arcsinh (- \frac{1 - x}{1 + x}) \Rightarrow$ $[F (x) = \arcsin (- \frac{1 - x}{1 + x}) has an uneven pole$ $at x = - 1, F^{'} (x) is increasing for - \infty < x < \infty$ $\bar{F} (x) = \arcsin (- \frac{1 - x}{∣ 1 + x ∣}) has an even pole$ $at x = - 1, {\bar{F}}^{'} (x) is increasing for - \infty < x < - 1$ $and decreasing for - 1 < x < \infty]$ $\Rightarrow 2 \int \frac{d x}{(1 + x) \sqrt{1 + x^{2}}} = \sqrt{2} arcsinh (- \frac{1 - x}{∣ 1 + x ∣})$
	Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jun/18
	$2)∫_0 ^1 (((√x) )/(1+x+x^2 +x^3 ))dx t^2 =x dx=2tdt ∫_0 ^1 ((t×2tdt)/(1+t^2 +t^4 +t^6 )) =2∫_0 ^1 (t^2 /((1+t^2 )+t^4 (1+t^2 )))dt =2∫_0 ^1 (t^2 /((1+t^2 )(1+t^4 )))dt =2∫_0 ^1 (((t^2 +t^4 )−(1+t^4 )+1)/((1+t^2 )(1+t^4 )))dt =2∫_0 ^1 (t^2 /(1+t^4 ))dt−2∫_0 ^1 (dt/(1+t^2 ))+2∫_0 ^1 (1/((1+t^2 )(1+t^4 )))dt =2∫_0 ^1 (dt/(t^2 +(1/t^2 )))−2∫_0 ^1 (dt/(1+t^2 ))+2∫_0 ^1 (dt/(t(t+(1/t))t^2 (t^2 +(1/t^2 )))) I_1 =∫_0 ^1 ((1−(1/t^2 )+1+(1/t^2 ))/(t^2 +(1/t^2 )))dt =∫_0 ^1 ((d(t+(1/t)))/((t+(1/t))^2 −2))+∫_0 ^1 ((d(t−(1/t)))/((t−(1/t))^2 +2)) now uzeformula ∫(dk/(k^2 −a^2 )) and ∫(dp/(p^2 +a^2 )) ={(1/(2(√2) ))ln∣((t+(1/t)−(√2) )/(t+(1/t)+(√2) ))∣+(1/(√2))tan^(−1) (((t−(1/t))/(√2)))}_0 ^1 =I_1 I_2 =−2∫_0 ^1 (dt/(1+t^2 )) ={−2×tan^(−1) (t)}_0 ^1 =I_2$
	$2) \int_{0}^{1} \frac{\sqrt{x}}{1 + x + x^{2} + x^{3}} d x$ $t^{2} = x d x = 2 t d t$ $\int_{0}^{1} \frac{t \times 2 t d t}{1 + t^{2} + t^{4} + t^{6}}$ $= 2 \int_{0}^{1} \frac{t^{2}}{(1 + t^{2}) + t^{4} (1 + t^{2})} d t$ $= 2 \int_{0}^{1} \frac{t^{2}}{(1 + t^{2}) (1 + t^{4})} d t$ $= 2 \int_{0}^{1} \frac{(t^{2} + t^{4}) - (1 + t^{4}) + 1}{(1 + t^{2}) (1 + t^{4})} d t$ $= 2 \int_{0}^{1} \frac{t^{2}}{1 + t^{4}} d t - 2 \int_{0}^{1} \frac{d t}{1 + t^{2}} + 2 \int_{0}^{1} \frac{1}{(1 + t^{2}) (1 + t^{4})} d t$ $= 2 \int_{0}^{1} \frac{d t}{t^{2} + \frac{1}{t^{2}}} - 2 \int_{0}^{1} \frac{d t}{1 + t^{2}} + 2 \int_{0}^{1} \frac{d t}{t (t + \frac{1}{t}) t^{2} (t^{2} + \frac{1}{t^{2}})}$ $I_{1} = \int_{0}^{1} \frac{1 - \frac{1}{t^{2}} + 1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} d t$ $= \int_{0}^{1} \frac{d (t + \frac{1}{t})}{{(t + \frac{1}{t})}^{2} - 2} + \int_{0}^{1} \frac{d (t - \frac{1}{t})}{{(t - \frac{1}{t})}^{2} + 2}$ $n o w u z e f o r m u l a \int \frac{d k}{k^{2} - a^{2}} a n d \int \frac{d p}{p^{2} + a^{2}}$ $= {\frac{1}{2 \sqrt{2}} l n ∣ \frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}} ∣ + \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}})}_{0}^{1} = I_{1}$ $I_{2} = - 2 \int_{0}^{1} \frac{d t}{1 + t^{2}}$ $= {- 2 \times {t a n}^{- 1} (t)}_{0}^{1} = I_{2}$
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