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Question Number 84021 by TANMAY PANACEA last updated on 08/Mar/20

Commented by abdomathmax last updated on 09/Mar/20
$I =∫_0 ^2 ((ln(1+2x))/(1+x^2 )) let f(a) =∫_0 ^2 ((ln(a+2x))/(1+x^2 ))dx we have f^′ (a) =(1/a)∫_0 ^2 (dx/((a+2x)(1+x^2 ))) let decompose F(x)=(1/((2x+a)(x^2 +1))) ⇒F(x)=(α/(2x+a)) +((βx+c)/(x^2 +1)) α =(1/(((−(a/2))^(2 ) +1))) =(1/((a^2 /4)+1)) =(4/(4+a^2 )) lim_(x→+∞) xF(x)=0=(α/2) +β ⇒β=−(α/2) =−(2/(4+a^2 )) F(0)=(1/a) =(α/a) +c ⇒c=(1/a)−(α/a) =((1−α)/a) =((1−(4/(4+a^2 )))/a) =(a^2 /(a(4+a^2 ))) ⇒ F(x)=(4/((a^2 +4)(2x+a))) +((((−2)/(a^2 +4))x +(a/(a^2 +4)))/(x^(2 ) +1)) =(1/(a^2 +4)){ (4/(2x+a)) +((−2x+a)/(x^2 +1))} ⇒ f^′ (a)=(1/(a(a^2 +4))){∫_0 ^2 ((4/(2x+a))+((−2x+a)/(x^2 +1)))dx} =(1/(a(a^2 +4))) ∫_0 ^2 ((2dx)/(x+(a/2))) −(2/(a(a^2 +4)))(1/2) ∫_0 ^2 ((2x−2a)/(x^2 +1))dx =(2/(a(a^2 +4)))[ln∣x+(a/2)∣]_0 ^2 −(1/(a(a^2 +4))) ∫_0 ^2 ((2xdx)/(x^2 +1)) +(1/((a^2 +4)))∫_0 ^2 (2/(x^2 +1))dx =(2/(a(a^2 +4)))(ln(2+(a/2))−ln((a/2)))−(1/(a(a^2 +4)))[ln(x^2 +1)]_0 ^2 +(2/(a^2 +4))[arctan(x)]_0 ^2 =(2/(a(a^2 +4)))ln(((4+a)/a))−((ln5)/(a(a^(2 ) +4))) +((2arctan(2))/(a^2 +4)) ⇒f(a)=∫_1 ^a (2/(t(t^2 +4)))ln(((t+4)/t))dt−ln(5)∫_1 ^a (dt/(t(t^2 +4))) +2arctan(2) ∫_1 ^a (dt/(t^2 +4)) +C ....be contonued...$
$I = \int_{0}^{2} \frac{l n (1 + 2 x)}{1 + x^{2}} l e t f (a) = \int_{0}^{2} \frac{l n (a + 2 x)}{1 + x^{2}} d x$ $w e h a v e f^{^{'}} (a) = \frac{1}{a} \int_{0}^{2} \frac{d x}{(a + 2 x) (1 + x^{2})} l e t d e c o m p o s e$ $F (x) = \frac{1}{(2 x + a) (x^{2} + 1)} \Rightarrow F (x) = \frac{α}{2 x + a} + \frac{β x + c}{x^{2} + 1}$ $α = \frac{1}{({(- \frac{a}{2})}^{2} + 1)} = \frac{1}{\frac{a^{2}}{4} + 1} = \frac{4}{4 + a^{2}}$ ${l i m}_{x \to + \infty} x F (x) = 0 = \frac{α}{2} + β \Rightarrow β = - \frac{α}{2} = - \frac{2}{4 + a^{2}}$ $F (0) = \frac{1}{a} = \frac{α}{a} + c \Rightarrow c = \frac{1}{a} - \frac{α}{a} = \frac{1 - α}{a}$ $= \frac{1 - \frac{4}{4 + a^{2}}}{a} = \frac{a^{2}}{a (4 + a^{2})} \Rightarrow$ $F (x) = \frac{4}{(a^{2} + 4) (2 x + a)} + \frac{\frac{- 2}{a^{2} + 4} x + \frac{a}{a^{2} + 4}}{x^{2} + 1}$ $= \frac{1}{a^{2} + 4} {\frac{4}{2 x + a} + \frac{- 2 x + a}{x^{2} + 1}} \Rightarrow$ $f^{^{'}} (a) = \frac{1}{a (a^{2} + 4)} {\int_{0}^{2} (\frac{4}{2 x + a} + \frac{- 2 x + a}{x^{2} + 1}) d x}$ $= \frac{1}{a (a^{2} + 4)} \int_{0}^{2} \frac{2 d x}{x + \frac{a}{2}} - \frac{2}{a (a^{2} + 4)} \frac{1}{2} \int_{0}^{2} \frac{2 x - 2 a}{x^{2} + 1} d x$ $= \frac{2}{a (a^{2} + 4)} {[l n ∣ x + \frac{a}{2} ∣]}_{0}^{2} - \frac{1}{a (a^{2} + 4)} \int_{0}^{2} \frac{2 x d x}{x^{2} + 1}$ $+ \frac{1}{(a^{2} + 4)} \int_{0}^{2} \frac{2}{x^{2} + 1} d x$ $= \frac{2}{a (a^{2} + 4)} (l n (2 + \frac{a}{2}) - l n (\frac{a}{2})) - \frac{1}{a (a^{2} + 4)} {[l n (x^{2} + 1)]}_{0}^{2}$ $+ \frac{2}{a^{2} + 4} {[a r c t a n (x)]}_{0}^{2}$ $= \frac{2}{a (a^{2} + 4)} l n (\frac{4 + a}{a}) - \frac{l n 5}{a (a^{2} + 4)} + \frac{2 a r c t a n (2)}{a^{2} + 4}$ $\Rightarrow f (a) = \int_{1}^{a} \frac{2}{t (t^{2} + 4)} l n (\frac{t + 4}{t}) d t - l n (5) \int_{1}^{a} \frac{d t}{t (t^{2} + 4)}$ $+ 2 a r c t a n (2) \int_{1}^{a} \frac{d t}{t^{2} + 4} + C . . . . b e c o n t o n u e d . . .$
	Answered by TANMAY PANACEA last updated on 08/Mar/20

	$1) \int_{1}^{\infty} \frac{2 x^{3} - 1}{x^{6} + 2 x^{3} + 9 x^{2} + 1} d x$ $= \int_{1}^{\infty} \frac{2 x^{3} - 1}{{(x^{3} + 1)}^{2} + 9 x^{2}} d x$ $= \int_{1}^{\infty} \frac{2 x - \frac{1}{x^{2}}}{{(\frac{x^{3} + 1}{x})}^{2} + 9} d x$ $\int_{1}^{\infty} \frac{2 x - \frac{1}{x^{2}}}{{(x^{2} + \frac{1}{x})}^{2} + 3^{2}} d x = \int_{1}^{\infty} \frac{d (x^{2} + \frac{1}{x})}{{(x^{2} + \frac{1}{x})}^{2} + 3^{2}}$ $\frac{1}{3} \times ∣ {t a n}^{- 1} (\frac{x^{2} + \frac{1}{x})}{3}) ∣_{1}^{\infty}$ $\frac{1}{3} \times ({t a n}^{- 1} \infty - {t a n}^{- 1} \frac{2}{3})$ $\frac{1}{3} \times (\frac{π}{2} - {t a n}^{- 1} \frac{2}{3})$
	Answered by TANMAY PANACEA last updated on 08/Mar/20

	Answered by TANMAY PANACEA last updated on 08/Mar/20

	$\int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{[{(s i n x)}^{2009} + {(c o s x)}^{2009}]}^{2}} {(s i n 2 x)}^{2008}$ $\int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{(s i n x)}^{4018} {[1 + {(c o t x)}^{2009}]}^{2}} \times 2^{2008} \times {s i n}^{2008} {x c o s}^{2008} x$ $\int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{[1 + {(c o t x)}^{2009}]}^{2}} \times 2^{2008} \times \frac{{c o s}^{2008} x}{{s i n}^{2010}} d x$ $\int_{0}^{\frac{π}{4}} \frac{l n (c o t x)}{{[1 + {(c o t x)}^{2009}]}^{2}} \times 2^{2008} \times {(c o t x)}^{2008} \times {c o s e c}^{2} x d x$ $t = c o t x \frac{d t}{d x} = - {c o z e c}^{2} x$ $\int_{\infty}^{1} \frac{l n t}{{(1 + t^{2009})}^{2}} \times 2^{2008} \times t^{2008} \times - d t$ $2^{2008} \int_{1}^{\infty} \frac{l n t}{{(1 + t^{2009})}^{2}} \times t^{2008} \times d t$
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