Math Question and Answers


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 55577 by rahul 19 last updated on 27/Feb/19

Commented by rahul 19 last updated on 27/Feb/19

$P r o v e t h a t :$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Feb/19
	$(dF/dα)=∫(∂/∂α)(((ln(1+αx))/(1+x^2 )))dx+((ln(1+α^2 ))/(1+α^2 ))×(dα/dα)−((ln(1+α×0))/(1+0^2 ))×(d/dα)(0) =∫_0 ^α (1/(1+x^2 ))×(1/(1+αx))×x dx+((ln(1+α^2 ))/(1+α^2 )) =∫_0 ^α (x/((1+x^2 )(1+αx)))dx+((ln(1+α^2 ))/(1+α^2 )) now (x/((1+x^2 )(1+αx)))=((ax+b)/(1+x^2 ))+(c/(1+αx)) x=(ax+b)(1+αx)+c+cx^2 x=ax+aαx^2 +b+bαx+c+cx^2 x=x^2 (aα+c)+x(a+bα)+b+c aα+c=0 a+bα=1 b+c=0 aα−b=0 a+bα=1 a+aα^2 =1 a=(1/(1+α^2 )) b=(α/(1+α^2 )) c=((−α)/(1+α^2 )) ∫_0 ^α (((ax+b)/(1+x^2 ))+(c/(1+αx)))dx (1/(2(1+α^2 )))∫_0 ^α ((2x)/(1+x^2 ))+(α/(1+α^2 ))∫_0 ^α (dx/(1+x^2 ))+((−α)/(1+α^2 ))∫(dx/(1+αx)) (1/(2(1+α^2 )))∫_0 ^α ((d(1+x^2 ))/(1+x^2 ))+(α/(1+α^2 ))∫_0 ^α (dx/(1+x^2 ))+((−α)/(1+α^2 ))∫_0 ^α (dx/(1+αx)) ∣(1/(2(1+α^2 )))ln(1+x^2 )+(α/(1+α^2 ))tan^(−1) (x)+((−α)/(1+α^2 ))×((ln(1+αx))/α)∣_0 ^α (1/(2(1+α^2 )))×ln(1+α^2 )+(α/(1+α^2 ))tan^(−1) (α)+((−1)/(1+α^2 ))×ln(1+α^2 ) =(α/(1+α^2 ))tan^(−1) (α)−(1/(2(1+α^2 )))×ln(1+α^2 ) so (dF/dα)=(α/(1+α^2 ))tan^(−1) (α)+((ln(1+α^2 ))/(2(1+α^2 ))) wait pls =tan^(−1) (α)(d/dα){ln(√(1+α^2 )) }+ln{(√(1+α^2 )) ×(d/dα)tan^(−1) α} =(d/dα){tan^(−1) α×ln(√(1+α^2 )) } so F=tan^(−1) (α)×ln(√(1+α^2 )) proved$
	$\frac{d F}{d α} = \int \frac{\partial}{\partial α} (\frac{l n (1 + α x)}{1 + x^{2}}) d x + \frac{l n (1 + α^{2})}{1 + α^{2}} \times \frac{d α}{d α} - \frac{l n (1 + α \times 0)}{1 + 0^{2}} \times \frac{d}{d α} (0)$ $= \int_{0}^{α} \frac{1}{1 + x^{2}} \times \frac{1}{1 + α x} \times x d x + \frac{l n (1 + α^{2})}{1 + α^{2}}$ $= \int_{0}^{α} \frac{x}{(1 + x^{2}) (1 + α x)} d x + \frac{l n (1 + α^{2})}{1 + α^{2}}$ $n o w$ $\frac{x}{(1 + x^{2}) (1 + α x)} = \frac{a x + b}{1 + x^{2}} + \frac{c}{1 + α x}$ $x = (a x + b) (1 + α x) + c + {c x}^{2}$ $x = a x + a α x^{2} + b + b α x + c + {c x}^{2}$ $x = x^{2} (a α + c) + x (a + b α) + b + c$ $a α + c = 0$ $a + b α = 1$ $b + c = 0$ $a α - b = 0$ $a + b α = 1$ $a + a α^{2} = 1$ $a = \frac{1}{1 + α^{2}} b = \frac{α}{1 + α^{2}} c = \frac{- α}{1 + α^{2}}$ $\int_{0}^{α} (\frac{a x + b}{1 + x^{2}} + \frac{c}{1 + α x}) d x$ $\frac{1}{2 (1 + α^{2})} \int_{0}^{α} \frac{2 x}{1 + x^{2}} + \frac{α}{1 + α^{2}} \int_{0}^{α} \frac{d x}{1 + x^{2}} + \frac{- α}{1 + α^{2}} \int \frac{d x}{1 + α x}$ $\frac{1}{2 (1 + α^{2})} \int_{0}^{α} \frac{d (1 + x^{2})}{1 + x^{2}} + \frac{α}{1 + α^{2}} \int_{0}^{α} \frac{d x}{1 + x^{2}} + \frac{- α}{1 + α^{2}} \int_{0}^{α} \frac{d x}{1 + α x}$ $∣ \frac{1}{2 (1 + α^{2})} l n (1 + x^{2}) + \frac{α}{1 + α^{2}} {t a n}^{- 1} (x) + \frac{- α}{1 + α^{2}} \times \frac{l n (1 + α x)}{α} ∣_{0}^{α}$ $\frac{1}{2 (1 + α^{2})} \times l n (1 + α^{2}) + \frac{α}{1 + α^{2}} {t a n}^{- 1} (α) + \frac{- 1}{1 + α^{2}} \times l n (1 + α^{2})$ $= \frac{α}{1 + α^{2}} {t a n}^{- 1} (α) - \frac{1}{2 (1 + α^{2})} \times l n (1 + α^{2})$ $s o$ $\frac{d F}{d α} = \frac{α}{1 + α^{2}} {t a n}^{- 1} (α) + \frac{l n (1 + α^{2})}{2 (1 + α^{2})}$ $w a i t p l s$ $= {t a n}^{- 1} (α) \frac{d}{d α} {l n \sqrt{1 + α^{2}}} + l n {\sqrt{1 + α^{2}} \times \frac{d}{d α} {t a n}^{- 1} α}$ $= \frac{d}{d α} {{t a n}^{- 1} α \times l n \sqrt{1 + α^{2}}}$ $s o F = {t a n}^{- 1} (α) \times l n \sqrt{1 + α^{2}} p r o v e d$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum