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Question Number 201553 by Calculusboy last updated on 08/Dec/23

	Answered by som(math1967) last updated on 09/Dec/23
	$1. ∫((1+logx−1)/((1+logx)^2 ))dx =∫(dx/((1+logx))) −∫(dx/((1+logx)^2 )) =(1/(1+logx))∫dx−∫{(d/dx)×(1/(1+logx))∫dx}dx −∫(dx/((1+logx)^2 )) =(x/(1+logx)) +∫((xdx)/(x(1+logx)^2 ))dx−∫(dx/((1+logx)^2 )) =(x/(1+logx)) +C$
	$1 . \int \frac{1 + l o g x - 1}{{(1 + l o g x)}^{2}} d x$ $= \int \frac{d x}{(1 + l o g x)} - \int \frac{d x}{{(1 + l o g x)}^{2}}$ $= \frac{1}{1 + l o g x} \int d x - \int {\frac{d}{d x} \times \frac{1}{1 + l o g x} \int d x} d x$ $- \int \frac{d x}{{(1 + l o g x)}^{2}}$ $= \frac{x}{1 + l o g x} + \int \frac{x d x}{x {(1 + l o g x)}^{2}} d x - \int \frac{d x}{{(1 + l o g x)}^{2}}$ $= \frac{x}{1 + l o g x} + C$
	Commented by Calculusboy last updated on 09/Dec/23

	$t h a n k s$
	Answered by som(math1967) last updated on 09/Dec/23
	$2.∫(e^((mtan^(−1) x)) /((1+x^2 )(√(1+x^2 ))))dx tan^(−1) x=t⇒(dx/(1+x^2 ))=dt (√(1+x^2 ))=(√(1+tan^2 t))=sect ∫((e^(mt) dt)/(sect)) I=∫coste^(mt) dt =cost∫e^(mt) dt−∫{(d/dt)cost∫e^(mt) dt}dt =((e^(mt) cost)/m) +(1/m)∫sinte^(mt) dt =((e^(mt) cost)/m^2 ) +((sint)/m)∫e^(mt) dt −(1/m)∫{(d/dt)sint∫e^(mt) dt}dt =((e^(mt) cost)/m^2 ) +((e^(mt) sint)/m^2 ) −(1/m^2 )∫e^(mt) costdt I=((e^(mt) (cost+sint))/m^2 )−(I/m^2 ) +C ⇒((I(m^2 +1))/m^2 ) =((e^(mt) (cost+sint))/m^2 ) +C ∴ I=((e^(mt) (cost+sint))/(m^2 +1)) +C_1 where t=tan^(−1) x$
	$2 . \int \frac{e^{(m \tan^{- 1} x)}}{(1 + x^{2}) \sqrt{1 + x^{2}}} d x$ $\tan^{- 1} x = t \Rightarrow \frac{d x}{1 + x^{2}} = d t$ $\sqrt{1 + x^{2}} = \sqrt{1 + {t a n}^{2} t} = s e c t$ $\int \frac{e^{m t} d t}{s e c t}$ $I = \int {c o s t e}^{m t} d t$ $= c o s t \int e^{m t} d t - \int {\frac{d}{d t} c o s t \int e^{m t} d t} d t$ $= \frac{e^{m t} c o s t}{m} + \frac{1}{m} \int {s i n t e}^{m t} d t$ $= \frac{e^{m t} c o s t}{m^{2}} + \frac{s i n t}{m} \int e^{m t} d t$ $- \frac{1}{m} \int {\frac{d}{d t} s i n t \int e^{m t} d t} d t$ $= \frac{e^{m t} c o s t}{m^{2}} + \frac{e^{m t} s i n t}{m^{2}} - \frac{1}{m^{2}} \int e^{m t} c o s t d t$ $I = \frac{e^{m t} (c o s t + s i n t)}{m^{2}} - \frac{I}{m^{2}} + C$ $\Rightarrow \frac{I (m^{2} + 1)}{m^{2}} = \frac{e^{m t} (c o s t + s i n t)}{m^{2}} + C$ $∴ I = \frac{e^{m t} (c o s t + s i n t)}{m^{2} + 1} + C_{1}$ $w h e r e t = \tan^{- 1} x$
	Commented by Calculusboy last updated on 09/Dec/23

	$t h a n k s$
	Answered by BaliramKumar last updated on 09/Dec/23

	$1 . put 1 + logx = t \Rightarrow dx = xdt$ $logx = t - 1$ $x = e^{t - 1}$ $\int \frac{(t - 1)}{t^{2}} \cdot e^{t - 1} dt$ $\frac{1}{e} \int e^{t} \cdot (\frac{1}{t} - \frac{1}{t^{2}}) dt \Rightarrow \int e^{x} [f (x) + f^{^{'}} (x)] = e^{x} \cdot f (x) + C$ $\frac{1}{e} \cdot e^{t} \cdot \frac{1}{t} + C = \frac{e^{t - 1}}{t} + C$ $\frac{x}{(1 + logx)} + C$
	Commented by Calculusboy last updated on 09/Dec/23

	$t h a n k s$
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