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Question Number 43657 by Tinkutara last updated on 13/Sep/18

Commented by maxmathsup by imad last updated on 24/Sep/18
$let A = ∫ (dx/((x^2 +2x +10)^2 )) we have A =∫ (dx/({(x+1)^2 +9}^2 )) changement x+1 =3tanθ give A = ∫ ((3 (1+tan^2 θ)dθ)/(81{1+tan^2 θ}^2 )) = (1/(27)) ∫ (dθ/(1+tan^2 θ)) =(1/(54)) ∫ (1+cos(2θ)dθ =(θ/(54)) +(1/(108)) sin(2θ) =(1/(54)) arctan(((x+1)/3)) +(1/(108)) ((2tanθ)/(1+tan^2 θ)) A=(1/(54))arctan(((x+1)/3)) +(1/(54)) ((x+1)/(3(1+(((x+1)/3))^2 ))) +c .$
$l e t A = \int \frac{d x}{{(x^{2} + 2 x + 10)}^{2}} w e h a v e A = \int \frac{d x}{{{(x + 1)}^{2} + 9}^{2}}$ $c h a n g e m e n t x + 1 = 3 t a n θ g i v e$ $A = \int \frac{3 (1 + {t a n}^{2} θ) d θ}{81 {1 + {t a n}^{2} θ}^{2}} = \frac{1}{27} \int \frac{d θ}{1 + {t a n}^{2} θ} = \frac{1}{54} \int (1 + c o s (2 θ) d θ$ $= \frac{θ}{54} + \frac{1}{108} s i n (2 θ) = \frac{1}{54} a r c t a n (\frac{x + 1}{3}) + \frac{1}{108} \frac{2 t a n θ}{1 + {t a n}^{2} θ}$ $A = \frac{1}{54} a r c t a n (\frac{x + 1}{3}) + \frac{1}{54} \frac{x + 1}{3 (1 + {(\frac{x + 1}{3})}^{2})} + c .$
	Answered by ajfour last updated on 13/Sep/18

	$\int \frac{d x}{{[{(x + 1)}^{2} + 3^{2}]}^{2}}$ $l e t x + 1 = 3 \tan θ$ $\Rightarrow \int \frac{3 \sec^{2} θ d θ}{81 \sec^{4} θ} = \frac{1}{54} \int 2 \cos^{2} θ d θ + c$ $. . . .$ $= \frac{1}{54} [\tan^{- 1} (\frac{x + 1}{3}) + \frac{3 (x + 1)}{(x^{2} + 2 x + 10)}] + c .$
	Commented by Tinkutara last updated on 14/Sep/18
	Thank you very much Sir! But in answer there is no multiplication with 3 in numerator of 2nd term.
	Commented by Tinkutara last updated on 14/Sep/18

	Commented by MJS last updated on 14/Sep/18

	$\frac{d}{d x} [\frac{1}{54} (\arctan \frac{x + 1}{3} + \frac{x + 1}{x^{2} + 2 x + 10}] =$ $= \frac{1}{54} (\frac{1}{3 (\frac{{(x + 1)}^{2}}{9} + 1)} + \frac{x^{2} + 2 x + 10 - (x + 1) (2 x + 2)}{{(x^{2} + 2 x + 10)}^{2}})$ $= \frac{1}{54} (\frac{3}{x^{2} + 2 x + 10} + \frac{- x^{2} - 2 x + 8}{{(x^{2} + 2 x + 10)}^{2}}) =$ $= \frac{1}{27} \times \frac{x^{2} + 2 x + 19}{{(x^{2} + 2 x + 10)}^{2}}$ $so the book is wrong$
	Commented by Tinkutara last updated on 14/Sep/18
	Thank you Sir!
	Answered by tanmay.chaudhury50@gmail.com last updated on 14/Sep/18
	$∫(dx/({(x+1)^2 +3^2 }^2 )) x+1=3tanα dx=3sec^2 αdα ∫((3sec^2 αdα)/({3^2 (1+tan^2 α)}^2 )) (1/(27))∫((sec^2 αdα)/(sec^4 α)) (1/(27))∫((1+cos2α)/2)dα (1/(54)){α+((sin2α)/2)}+c (1/(54)){tan^(−1) (((x+1)/3))+(1/2)×((2tanα)/(1+tan^2 α))} (1/(54)){tan^(−1) (((x+1)/3))+(((x+1)/3)/(1+(((x+1)/3))^2 ))}+c (1/(54)){tan^(−1) (((x+1)/3))+((3(x+1))/(x^2 +2x+10))}+c$
	$\int \frac{d x}{{{(x + 1)}^{2} + 3^{2}}^{2}}$ $x + 1 = 3 t a n α$ $d x = 3 {s e c}^{2} α d α$ $\int \frac{3 {s e c}^{2} α d α}{{3^{2} (1 + {t a n}^{2} α)}^{2}}$ $\frac{1}{27} \int \frac{{s e c}^{2} α d α}{{s e c}^{4} α}$ $\frac{1}{27} \int \frac{1 + c o s 2 α}{2} d α$ $\frac{1}{54} {α + \frac{s i n 2 α}{2}} + c$ $\frac{1}{54} {{t a n}^{- 1} (\frac{x + 1}{3}) + \frac{1}{2} \times \frac{2 t a n α}{1 + {t a n}^{2} α}}$ $\frac{1}{54} {{t a n}^{- 1} (\frac{x + 1}{3}) + \frac{\frac{x + 1}{3}}{1 + {(\frac{x + 1}{3})}^{2}}} + c$ $\frac{1}{54} {{t a n}^{- 1} (\frac{x + 1}{3}) + \frac{3 (x + 1)}{x^{2} + 2 x + 10}} + c$
	Commented by Tinkutara last updated on 14/Sep/18
	Thanks very much Sir! But please check the answer in book.
	Answered by MJS last updated on 14/Sep/18

	$\int \frac{d x}{{(x^{2} + 2 x + 10)}^{2}} = \int \frac{d x}{{({(x + 1)}^{2} + 9)}^{2}} =$ $[t = x + 1 \to d x = d t]$ $= \int \frac{d t}{{(t^{2} + 9)}^{2}} =$ $[reduction formula$ $\int \frac{d t}{{({a t}^{2} + b)}^{n}} = \frac{t}{2 b (n - 1) {({a t}^{2} + b)}^{n - 1}} + \frac{2 n - 3}{2 b (n - 1)} \int \frac{d t}{{({a t}^{2} + b)}^{n - 1}}]$ $= \frac{t}{18 (t^{2} + 9)} + \frac{1}{18} \int \frac{d t}{t^{2} + 9} =$ $\int \frac{d t}{t^{2} + 9} =$ $[u = \frac{t}{3} \to d t = 3 d u]$ $= \frac{1}{3} \int \frac{d u}{u^{2} + 1} = \frac{1}{3} \arctan u = \frac{1}{3} \arctan \frac{t}{3}$ $= \frac{t}{18 (t^{2} + 9)} + \frac{1}{54} \arctan \frac{t}{3} =$ $= \frac{x + 1}{18 (x^{2} + 2 x + 10)} + \frac{1}{54} \arctan \frac{x + 1}{3} + C$
	Commented by Tinkutara last updated on 14/Sep/18
	Thanks very much.
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