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 41555 by Raj Singh last updated on 09/Aug/18

	Answered by alex041103 last updated on 09/Aug/18

	$1 . \int \frac{l n x}{{(1 + l n x)}^{2}} d x$ $l e t x = e^{u} d x = e^{u} d u .$ $\int \frac{1 + u - 1}{{(1 + u)}^{2}} e^{u} d u = \int \frac{e^{u}}{1 + u} d u - \int \frac{e^{u}}{{(1 + u)}^{2}} d u =$ $= \int \frac{e^{u}}{1 + u} d u + \int e^{u} d (\frac{1}{1 + u}) = n o w b y p a r t s$ $= \int \frac{e^{u}}{1 + u} d u + \frac{e^{u}}{1 + u} - \int \frac{e^{u}}{1 + u} d u = \frac{e^{u}}{1 + u}$ $\Rightarrow \int \frac{l n x}{{(1 + l n x)}^{2}} d x = \frac{x}{1 + l n x} + C$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18

	$1) \int \frac{1 + l n x - 1}{{(1 + l n x)}^{2}} d x$ $\int \frac{(1 + l n x) \frac{d x}{d x} - x . \frac{d}{d x} (1 + l n x)}{{(1 + l n x)}^{2}} d x$ $\int \frac{d}{d x} (\frac{x}{1 + l n x}) d x$ $\int d (\frac{x}{1 + l n x})$ $\frac{x}{1 + l n x} + c$
	Answered by alex041103 last updated on 09/Aug/18

	$2 . \int \frac{s i n (l n x)}{x^{3}} d x$ $F i r s t w a y : s i m p l y i n t e g r a t e$ $x = e^{u} d x = e^{u} d u$ $\int \frac{s i n (l n x)}{x^{3}} d x = \int \frac{s i n (u)}{e^{3 u}} e^{u} d u =$ $= \int e^{- 2 u} s i n (u) d u = I$ $I B P (1)$ $I = - \frac{1}{2} e^{- 2 u} s i n (u) + \frac{1}{2} \int e^{- 2 u} c o s (u) d u$ $I B P (2)$ $I = - \frac{1}{2} e^{- 2 u} s i n (u) + \frac{1}{2} (- \frac{1}{2} e^{- 2 u} c o s (u) - \frac{1}{2} \int e^{- 2 u} s i n (u) d u) =$ $= - \frac{1}{2} e^{- 2 u} s i n (u) - \frac{1}{4} e^{- 2 u} c o s (u) - \frac{1}{4} I$ $\Rightarrow \frac{5}{4} I = - \frac{1}{2} e^{- 2 u} s i n (u) - \frac{1}{4} e^{- 2 u} c o s (u)$ $\Rightarrow \int e^{- 2 u} s i n (u) d u = - \frac{2}{5} e^{- 2 u} s i n (u) - \frac{1}{5} e^{- 2 u} c o s (u)$ $\int e^{- 2 u} s i n (u) d u = - \frac{1}{5} e^{- 2 u} (2 s i n (u) + c o s (u))$ $\Rightarrow \int \frac{s i n (l n x)}{x^{3}} d x = - \frac{1}{5 x^{2}} (2 s i n (l n x) + c o s (l n x)) + C$
	Commented by alex041103 last updated on 09/Aug/18

	$S e c o n d w a y : C o m p l e x A n a l y s i s$ $s i n (l n x) = I m (e^{i l n x}) = I m (x^{i})$ $\Rightarrow \int \frac{s i n (l n x)}{x^{3}} d x = I m (\int x^{i - 3} d x) =$ $= I m (\frac{x^{i - 2}}{i - 2} \frac{- i - 2}{- i - 2}) = - \frac{1}{5 x^{2}} I m ((2 + i) x^{i}) =$ $= - \frac{1}{5 x^{2}} I m ((2 + i) (c o s (l n x) + i s i n (l n x))) =$ $= - \frac{1}{5 x^{2}} (2 s i n (l n x) + c o s (l n x))$ $\Rightarrow \int \frac{s i n (l n x)}{x^{3}} d x = - \frac{1}{5 x^{2}} (2 s i n (l n x) + c o s (l n x)) + C$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18
	$2)∫((sin(lnx))/x^3 )dx t=lnx dt=(dx/x) [and x=e^t ∫((sint)/e^(2t) )dt ∫e^(−2t) sint dt p=∫e^(−2t) cost dt q=∫e^(−2t) sint dt p+iq=∫e^(−2t) .e^(it) dt p+iq=∫e^(t(−2+i)) dt =(e^(t(−2+i)) /(−2+i))+c =((e^(−2t) .e^(it) )/((−2+i)(−2−i)))×(−2−i)+c =e^(−2t) .(((cost+isint))/(4+1))×(−2−i)+c =(e^(−2t) /5)(cost+isint)(−2−i)+c =(e^(−2t) /5)(−2cost−icost−i2sint+sint)+c =(e^(−2t) /5){(−2cost+sint)+i(−cost−2sint)}+c real part =(e^(−2t) /5)(−2cost+sint)+i(e^(−2t) /5)(−cost−2sint)+c ∫e^(−2t) sint dt =(e^(−2t) /5)(−cost−2sint)+c =(x^(−2) /5){−cos(lnx)−2sin(lnx)}+c$
	$2) \int \frac{s i n (l n x)}{x^{3}} d x$ $t = l n x d t = \frac{d x}{x} [a n d x = e^{t}$ $\int \frac{s i n t}{e^{2 t}} d t$ $\int e^{- 2 t} s i n t d t$ $p = \int e^{- 2 t} c o s t d t$ $q = \int e^{- 2 t} s i n t d t$ $p + i q = \int e^{- 2 t} . e^{i t} d t$ $p + i q = \int e^{t (- 2 + i)} d t$ $= \frac{e^{t (- 2 + i)}}{- 2 + i} + c$ $= \frac{e^{- 2 t} . e^{i t}}{(- 2 + i) (- 2 - i)} \times (- 2 - i) + c$ $= e^{- 2 t} . \frac{(c o s t + i s i n t)}{4 + 1} \times (- 2 - i) + c$ $= \frac{e^{- 2 t}}{5} (c o s t + i s i n t) (- 2 - i) + c$ $= \frac{e^{- 2 t}}{5} (- 2 c o s t - i c o s t - i 2 s i n t + s i n t) + c$ $= \frac{e^{- 2 t}}{5} {(- 2 c o s t + s i n t) + i (- c o s t - 2 s i n t)} + c$ $r e a l p a r t$ $= \frac{e^{- 2 t}}{5} (- 2 c o s t + s i n t) + i \frac{e^{- 2 t}}{5} (- c o s t - 2 s i n t) + c$ $\int e^{- 2 t} s i n t d t$ $= \frac{e^{- 2 t}}{5} (- c o s t - 2 s i n t) + c$ $= \frac{x^{- 2}}{5} {- c o s (l n x) - 2 s i n (l n x)} + c$
	Answered by maxmathsup by imad last updated on 09/Aug/18
	$let I = ∫ ((sin(ln(x)))/x^3 ) dx chamgement ln(x) =t give I = ∫ ((sin(t))/e^(3t) ) e^t dt =∫ e^(−2t) sint dt = Im( ∫ e^(−2t +it) dt) but ∫ e^((−2+i)t) dt = (1/(−2+i)) e^((−2+i)t) +c =−((2+i)/5) e^(−2t) (cost +isnt) +c =−(1/5) e^(−2t) { 2cost +2isint +icost −sint}+c =−(1/5) e^(−2t) {2cost −sint +i(2sint +cost)} ⇒ I =−(e^(−2t) /5){ 2sin(t)+cos(t)} +k$
	$l e t I = \int \frac{s i n (l n (x))}{x^{3}} d x c h a m g e m e n t l n (x) = t g i v e$ $I = \int \frac{s i n (t)}{e^{3 t}} e^{t} d t = \int e^{- 2 t} s i n t d t = I m (\int e^{- 2 t + i t} d t) b u t$ $\int e^{(- 2 + i) t} d t = \frac{1}{- 2 + i} e^{(- 2 + i) t} + c = - \frac{2 + i}{5} e^{- 2 t} (c o s t + i s n t) + c$ $= - \frac{1}{5} e^{- 2 t} {2 c o s t + 2 i s i n t + i c o s t - s i n t} + c$ $= - \frac{1}{5} e^{- 2 t} {2 c o s t - s i n t + i (2 s i n t + c o s t)} \Rightarrow$ $I = - \frac{e^{- 2 t}}{5} {2 s i n (t) + c o s (t)} + k$
	Commented by maxmathsup by imad last updated on 09/Aug/18
	$but t =ln(x) ⇒ I =−(1/(5x^2 )){ 2sin(ln(x)) +cos(ln(x))} +k$
	$b u t t = l n (x) \Rightarrow I = - \frac{1}{5 x^{2}} {2 s i n (l n (x)) + c o s (l n (x))} + k$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum