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Question Number 66938 by Cmr 237 last updated on 20/Aug/19

Commented by mathmax by abdo last updated on 21/Aug/19

$8) b y p a r t s \int l n (1 + x^{2}) d x = x l n (1 + x^{2}) - \int x \frac{2 x}{1 + x^{2}} d x$ $= x l n (1 + x^{2}) - 2 \int \frac{1 + x^{2} - 1}{1 + x^{2}} d x = x l n (1 + x^{2}) - 2 x + 2 a r c t a n x + C$ $7) b y p a r t s \int a r c t a n x d x = x a r c t a n (x) - \int \frac{x}{1 + x^{2}} d x$ $= x a r c t a n (x) - \frac{1}{2} l n (1 + x^{2}) + C$
Commented by mathmax by abdo last updated on 21/Aug/19

$6) w e d o t h e c h a n g e m e n t a r c s i n x = t \Rightarrow x = s i n t \Rightarrow$ $\int a r c s i n x d x = \int t c o s t d t =_{b y p a r t s} t s i n t - \int s i n t d t$ $= t s i n t + c o s t + C = x a r c s i n x + \sqrt{1 - x^{2}} + C .$
Commented by mathmax by abdo last updated on 21/Aug/19

$5) w e d o t h e c h a n g e m e n t t a n (\frac{x}{2}) = t \Rightarrow$ $\int \frac{d x}{s i n x} = \int \frac{1}{\frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{d t}{t} = l n ∣ t ∣ + C = l n ∣ t a n (\frac{x}{2}) ∣ + C$
Commented by mathmax by abdo last updated on 21/Aug/19
$4) let I =∫ (dx/(x(√(x^2 +x+1)))) we have x^(2 ) +x+1 =(x+(1/2))^2 +(3/4) so we do the changement x+(1/2) =((√3)/2) sh(t) ⇒ I =∫ (1/((((√3)/2)sh(t)−(1/2))((√3)/2)ch(t)))((√3)/2)ch(t)dt =∫ ((2dt)/((√3)sh(t)−1)) =∫((2dt)/((√3)(((e^t −e^(−t) )/2))−1)) =∫ ((4dt)/((√3)e^t −(√3)e^(−t) −2)) =_(e^t =u) ∫ (4/((√3)u−(√3)u^(−1) −2))(du/u) =∫ ((4du)/((√3)u^2 −(√3)−2u)) let decompose F(u)=(4/((√3)u^2 −2u−(√3))) (√3)u^2 −2u−(√3)=0→ Δ^′ =1+3 =4 ⇒u_1 =((1+2)/(√3)) =(√3) and u_2 =((1−2)/(√3)) =−(1/(√3)) ⇒F(u) =(4/((√3)(u−(√3))(u+(1/(√3))))) ⇒ I =(4/((√3)((√3)+(1/(√3)))))∫ {(1/(u−(√3)))−(1/(u+(1/(√3))))}du =ln∣((u−(√3))/(u+(1/(√3))))∣ +C =ln∣((e^t −(√3))/(e^t +(1/(√3))))∣ +C but sh(t)=((2x+1)/(√3)) ⇒t =argsh(((2x+1)/(√3))) =ln(((2x+1)/(√3))+(√(1+(((2x+1)/((√3) )))^2 ))) ⇒e^t =((2x+1)/(√3))+(√(1+(((2x+1)/(√3)))^2 )) ⇒ I =ln∣((2x+1)/(√3))+(√(1+(((2x+1)/(√3)))^2 ))−(√3)∣−ln∣((2x+1)/(√3))+(√(1+(((2x+1)/(√3)))^2 ))+(1/(√3))∣+C$
$4) l e t I = \int \frac{d x}{x \sqrt{x^{2} + x + 1}} w e h a v e x^{2} + x + 1 = {(x + \frac{1}{2})}^{2} + \frac{3}{4}$ $s o w e d o t h e c h a n g e m e n t x + \frac{1}{2} = \frac{\sqrt{3}}{2} s h (t) \Rightarrow$ $I = \int \frac{1}{(\frac{\sqrt{3}}{2} s h (t) - \frac{1}{2}) \frac{\sqrt{3}}{2} c h (t)} \frac{\sqrt{3}}{2} c h (t) d t = \int \frac{2 d t}{\sqrt{3} s h (t) - 1}$ $= \int \frac{2 d t}{\sqrt{3} (\frac{e^{t} - e^{- t}}{2}) - 1} = \int \frac{4 d t}{\sqrt{3} e^{t} - \sqrt{3} e^{- t} - 2} =_{e^{t} = u} \int \frac{4}{\sqrt{3} u - \sqrt{3} u^{- 1} - 2} \frac{d u}{u}$ $= \int \frac{4 d u}{\sqrt{3} u^{2} - \sqrt{3} - 2 u} l e t d e c o m p o s e F (u) = \frac{4}{\sqrt{3} u^{2} - 2 u - \sqrt{3}}$ $\sqrt{3} u^{2} - 2 u - \sqrt{3} = 0 \to Δ^{^{'}} = 1 + 3 = 4 \Rightarrow u_{1} = \frac{1 + 2}{\sqrt{3}} = \sqrt{3}$ $a n d u_{2} = \frac{1 - 2}{\sqrt{3}} = - \frac{1}{\sqrt{3}} \Rightarrow F (u) = \frac{4}{\sqrt{3} (u - \sqrt{3}) (u + \frac{1}{\sqrt{3}})} \Rightarrow$ $I = \frac{4}{\sqrt{3} (\sqrt{3} + \frac{1}{\sqrt{3}})} \int {\frac{1}{u - \sqrt{3}} - \frac{1}{u + \frac{1}{\sqrt{3}}}} d u = l n ∣ \frac{u - \sqrt{3}}{u + \frac{1}{\sqrt{3}}} ∣ + C$ $= l n ∣ \frac{e^{t} - \sqrt{3}}{e^{t} + \frac{1}{\sqrt{3}}} ∣ + C b u t s h (t) = \frac{2 x + 1}{\sqrt{3}} \Rightarrow t = a r g s h (\frac{2 x + 1}{\sqrt{3}})$ $= l n (\frac{2 x + 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x + 1}{\sqrt{3}})}^{2}}) \Rightarrow e^{t} = \frac{2 x + 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x + 1}{\sqrt{3}})}^{2}}$ $\Rightarrow I = l n ∣ \frac{2 x + 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x + 1}{\sqrt{3}})}^{2}} - \sqrt{3} ∣ - l n ∣ \frac{2 x + 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x + 1}{\sqrt{3}})}^{2}} + \frac{1}{\sqrt{3}} ∣ + C$
Commented by Cmr 237 last updated on 21/Aug/19

$thank sir (merci beaucoup monsieur)$
Commented by mathmax by abdo last updated on 21/Aug/19

$y o u a r e w e l c o m e s i r .$
Commented by mathmax by abdo last updated on 22/Aug/19

$3) \int {c o s}^{4} x d x = \int {(\frac{1 + c o s (2 x)}{2})}^{2} d x = \frac{1}{4} \int (1 + 2 c o s (2 x) + {c o s}^{2} (2 x)) d x$ $= \frac{1}{4} x + \frac{1}{2} \int c o s (2 x) d x + \frac{1}{8} \int (1 + c o s (4 x)) d x$ $= \frac{x}{4} + \frac{1}{4} s i n (2 x) + \frac{x}{8} + \frac{1}{32} s i n (4 x) + C$
	Answered by Kunal12588 last updated on 21/Aug/19

	$1) I = \int t a n h x d x$ $I = \int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x$ $l e t e^{x} + e^{- x} = t$ $\Rightarrow (e^{x} - e^{- x}) d x = d t$ $\Rightarrow I = \int \frac{d t}{t}$ $\Rightarrow I = l n t + C$ $\Rightarrow I = l n ∣ e^{x} + e^{- x} ∣ + c$ $\Rightarrow I = l n ∣ c o s h x ∣ + C$
	Answered by Kunal12588 last updated on 21/Aug/19

	$2) I = \int \frac{x}{1 + x^{4}} d x$ $l e t x^{2} = t$ $\Rightarrow x d x = \frac{d t}{2}$ $\Rightarrow I = \frac{1}{2} \int \frac{d t}{1 + t^{2}}$ $\Rightarrow I = \frac{1}{2} {t a n}^{- 1} t + C$ $\Rightarrow I = \frac{1}{2} {t a n}^{- 1} x^{2} + C$
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