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Question Number 85896 by ar247 last updated on 25/Mar/20

Commented by abdomathmax last updated on 25/Mar/20

$I = \int \frac{5 x - 5}{3 x^{2} - 8 x - 3} d x$ $3 x^{2} - 8 x - 3 = 0 \to Δ^{^{'}} = 4^{2} - (3) (- 3) = 16 + 9 = 25$ $x_{1} = \frac{4 + 5}{3} = 3 a n d x_{2} = \frac{4 - 5}{3} = - \frac{1}{3} \Rightarrow$ $I = \int \frac{5 x - 5}{3 (x - 3) (x + \frac{1}{3})} d x l e t d e c o m p o s e$ $F (x) = \frac{5 x - 5}{(x - 3) (3 x + 1)}$ $F (x) = \frac{a}{x - 3} + \frac{b}{3 x + 1}$ $a = (x - 3) F (x) ∣_{x = 3} = \frac{10}{10} = 1$ $b = (3 x + 1) F (x) ∣_{x = - \frac{1}{3}} = \frac{- \frac{5}{3} - 5}{- \frac{1}{3} - 3} = \frac{20}{10} = 2 \Rightarrow$ $F (x) = \frac{1}{x - 3} + \frac{2}{3 x + 1} \Rightarrow I = \int \frac{d x}{x - 3} + 2 \int \frac{d x}{3 x + 1}$ $= l n ∣ x - 3 ∣ + \frac{2}{3} l n ∣ 3 x + 1 ∣ + c$
Commented by abdomathmax last updated on 25/Mar/20

$2) A = \int \frac{2 x^{5} - x^{3} - 1}{x^{3} - 4 x} d x \Rightarrow$ $A = \int \frac{2 (x^{3} - 4 x) x^{2} + 8 x^{3} - x^{3} - 1}{x^{3} - 4 x} d x$ $= 2 x + \int \frac{7 x^{3} - 1}{x^{3} - 4 x} d x$ $= 2 x + \int \frac{7 (x^{3} - 4 x) + 28 x - 1}{x^{3} - 4 x} d x$ $= 2 x + 7 + \int \frac{28 x - 1}{x^{3} - 4 x} d x l e t d e c o m p o s e$ $F (x) = \frac{28 x - 1}{x^{3} - 4 x} = \frac{28 x - 1}{x (x - 2) (x + 2)}$ $F (x) = \frac{a}{x} + \frac{b}{x - 2} + \frac{c}{x + 2}$ $a = \frac{- 1}{(- 2) (2)} = \frac{1}{4}$ $b = \frac{56 - 1}{2.4} = \frac{55}{8}$ $c = \frac{- 56 - 1}{(- 2) (- 4)} = \frac{57}{8} \Rightarrow F (x) = \frac{1}{4 x} + \frac{55}{8 (x - 2)} + \frac{57}{8 (x + 2)}$ $A = 2 x + 7 + \frac{1}{4} l n ∣ x ∣ + \frac{55}{8} l n ∣ x - 2 ∣ + \frac{57}{8} l n ∣ x + 2 ∣ + C$
Commented by abdomathmax last updated on 25/Mar/20
$4) I = ∫ ((√(x^2 −25))/x)dx we do the chsngement x=5 ch(t) ⇒ I =∫ ((5 sh(t))/(5ch(t))) ×5sh(t)dt =5 ∫ ((sh^2 t)/(cht)) dt =(5/2)∫ ((ch(2t)−1)/(ch(t)))dt =(5/2)∫ ((((e^(2t) +e^(−2t) )/2)−1)/((e^t +e^(−t) )/2))dt =(5/2)∫ ((e^(2t) +e^(−2t) −2)/(e^t +e^(−t) ))dt =_(e^t =u) (5/2)∫ ((u^2 +u^(−2) −2)/(u+u^(−1) ))(du/u) =(5/2)∫ ((u^2 +u^(−2) −2)/(u^2 +1))du =(5/2)∫ ((u^4 +1−2u^2 )/(u^2 (u^2 +1)))du decomposition ((u^4 −2u^2 +1)/(u^4 +u^2 )) =((u^4 +u^2 −3u^2 +1)/(u^4 +u^2 )) =1 +((−3u^2 +1)/(u^4 +u^2 )) F(u) =((−3u^2 +1)/(u^4 +u^2 )) =((−3u^2 +1)/(u^2 (u^2 +1))) =(a/u) +(b/u^2 ) +((cu +d)/(u^2 +1)) F(−u)=F(u) ⇒((−a)/u) +(b/u^2 ) +((−cu +d)/(u^2 +1))=(a/u) +(b/u^2 )+((cu+d)/(u^2 +1)) ⇒a=0 and c=0 ⇒F(u)=(b/u^2 ) +(d/(u^2 +1)) b=1 ⇒F(u)=(1/u^2 ) +(d/(u^2 +1)) F(1)=−1 =1 +(d/2) ⇒−2 =(d/2) ⇒d=−4 ⇒ F(u)=(1/u^2 )−(4/(u^2 +1)) ⇒ I =(5/2){u +∫ F(u)du} =(5/2)u +(5/2)(−(1/u)−4 arctan(u)) +C =(5/2)e^t −(5/2)e^(−t) −10 arctan(e^t )+C t=argch((x/5)) =ln((x/5)+(√((x^2 /(25))−1))) ⇒ I =(5/2){ (x/5)+(√((x^2 /(25))−1))−(1/((x/5)+(√((x^2 /(25))−1))))} −10 arctan((x/5)+(√((x^2 /(25))−1))) +C$
$4) I = \int \frac{\sqrt{x^{2} - 25}}{x} d x w e d o t h e c h s n g e m e n t$ $x = 5 c h (t) \Rightarrow I = \int \frac{5 s h (t)}{5 c h (t)} \times 5 s h (t) d t$ $= 5 \int \frac{{s h}^{2} t}{c h t} d t = \frac{5}{2} \int \frac{c h (2 t) - 1}{c h (t)} d t$ $= \frac{5}{2} \int \frac{\frac{e^{2 t} + e^{- 2 t}}{2} - 1}{\frac{e^{t} + e^{- t}}{2}} d t = \frac{5}{2} \int \frac{e^{2 t} + e^{- 2 t} - 2}{e^{t} + e^{- t}} d t$ $=_{e^{t} = u} \frac{5}{2} \int \frac{u^{2} + u^{- 2} - 2}{u + u^{- 1}} \frac{d u}{u}$ $= \frac{5}{2} \int \frac{u^{2} + u^{- 2} - 2}{u^{2} + 1} d u$ $= \frac{5}{2} \int \frac{u^{4} + 1 - 2 u^{2}}{u^{2} (u^{2} + 1)} d u d e c o m p o s i t i o n$ $\frac{u^{4} - 2 u^{2} + 1}{u^{4} + u^{2}} = \frac{u^{4} + u^{2} - 3 u^{2} + 1}{u^{4} + u^{2}}$ $= 1 + \frac{- 3 u^{2} + 1}{u^{4} + u^{2}}$ $F (u) = \frac{- 3 u^{2} + 1}{u^{4} + u^{2}} = \frac{- 3 u^{2} + 1}{u^{2} (u^{2} + 1)}$ $= \frac{a}{u} + \frac{b}{u^{2}} + \frac{c u + d}{u^{2} + 1}$ $F (- u) = F (u) \Rightarrow \frac{- a}{u} + \frac{b}{u^{2}} + \frac{- c u + d}{u^{2} + 1} = \frac{a}{u} + \frac{b}{u^{2}} + \frac{c u + d}{u^{2} + 1}$ $\Rightarrow a = 0 a n d c = 0 \Rightarrow F (u) = \frac{b}{u^{2}} + \frac{d}{u^{2} + 1}$ $b = 1 \Rightarrow F (u) = \frac{1}{u^{2}} + \frac{d}{u^{2} + 1}$ $F (1) = - 1 = 1 + \frac{d}{2} \Rightarrow - 2 = \frac{d}{2} \Rightarrow d = - 4 \Rightarrow$ $F (u) = \frac{1}{u^{2}} - \frac{4}{u^{2} + 1} \Rightarrow$ $I = \frac{5}{2} {u + \int F (u) d u}$ $= \frac{5}{2} u + \frac{5}{2} (- \frac{1}{u} - 4 a r c t a n (u)) + C$ $= \frac{5}{2} e^{t} - \frac{5}{2} e^{- t} - 10 a r c t a n (e^{t}) + C$ $t = a r g c h (\frac{x}{5}) = l n (\frac{x}{5} + \sqrt{\frac{x^{2}}{25} - 1}) \Rightarrow$ $I = \frac{5}{2} {\frac{x}{5} + \sqrt{\frac{x^{2}}{25} - 1} - \frac{1}{\frac{x}{5} + \sqrt{\frac{x^{2}}{25} - 1}}}$ $- 10 a r c t a n (\frac{x}{5} + \sqrt{\frac{x^{2}}{25} - 1}) + C$
Commented by Kunal12588 last updated on 26/Mar/20

$s i r i n (2) \int 2 x^{2} d x = \frac{2}{3} x^{3} ?$
	Answered by jagoll last updated on 26/Mar/20

	$(3) \int \frac{dx}{\sqrt{x^{2} + 4}} ?$ $[x = 2 \tan u]$ $\int \frac{2 \sec^{2} u}{2 \sec u} du = \int \sec u du$ $= \ln ∣ \sec u + \tan u ∣ + c$ $= \ln ∣ \frac{x + \sqrt{x^{2} + 4}}{2} ∣ + c$
	Commented by Kunal12588 last updated on 26/Mar/20

	$= l n ∣ x + \sqrt{x^{2} + 4} ∣ + C$
	Commented by john santu last updated on 26/Mar/20

	$s a m e i t s i r$ $\ln ∣ \frac{x + \sqrt{x^{2} + 4}}{2} ∣ + c =$ $\ln ∣ x + \sqrt{x^{2} + 4} ∣ + C$
	Answered by Kunal12588 last updated on 26/Mar/20

	$(5) I = \int \frac{d x}{\sqrt{5 - 4 x - 2 x^{2}}}$ $= \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{- (x^{2} + 2 x - \frac{5}{2})}}$ $= \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{- [{(x + 1)}^{2} - \frac{7}{2}]}}$ $= \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{7}{2} - {(x + 1)}^{2}}}$ $= \frac{1}{\sqrt{2}} {s i n}^{- 1} (\frac{x + 1}{\frac{\sqrt{7}}{\sqrt{2}}}) + C$ $= \frac{1}{\sqrt{2}} {s i n}^{- 1} (\frac{\sqrt{2} x + \sqrt{2}}{\sqrt{7}}) + C$
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