∫ ((e^(2x) +1)/(2e^x −1)) dx = ?


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Question Number 35920 by rahul 19 last updated on 25/May/18

$\int \frac{e^{2 x} + 1}{2 e^{x} - 1} d x = ?$
Commented by prof Abdo imad last updated on 26/May/18

$c h a n g e m e n t e^{x} = t g i v e$ $I = \int \frac{t^{2} + 1}{2 t - 1} \frac{d t}{t} = \int \frac{t^{2} + 1}{2 t^{2} - t} d t$ $= \frac{1}{2} \int \frac{2 t^{2} + 2 - t + t}{2 t^{2} - t} d t$ $= \frac{t}{2} + \frac{1}{2} \int \frac{2 + t}{2 t^{2} - t} d t l e t d e c o m p o s e$ $F (t) = \frac{2 + t}{t (2 t - 1)} = \frac{a}{t} + \frac{b t + c}{2 t - 1}$ $\Leftrightarrow 2 a t - a + {b t}^{2} + c t = 2 + t \Leftrightarrow {b t}^{2} + (2 a + c) t - a$ $= 2 + t \Rightarrow b = 0, a = - 2, 2 a + c = 1 \Rightarrow c = 1 - 2 (- 2) = 5$ $F (t) = \frac{- 2}{t} + \frac{5}{2 t - 1}$ $\int \frac{2 + t}{2 t^{2} - t} d t = - 2 l n ∣ t ∣ + \frac{5}{2} l n ∣ t - \frac{1}{2} ∣ + c \Rightarrow$ $I = \frac{t}{2} - l n ∣ t ∣ + \frac{5}{4} l n ∣ t - \frac{1}{2} ∣ + c \Rightarrow$ $I = \frac{e^{x}}{2} - x + \frac{5}{4} l n ∣ e^{x} - \frac{1}{2} ∣ + c .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/May/18
	$t=2e^x −1 e^x =((t+1)/2) so e^x dx=(1/2)dt dx=(1/2)×(2/(t+1))×dt ∫(((((t+1)/2))^2 +1)/t)×(dt/(t+1)) ∫((t^2 +2t+1+4)/(4t(t+1)))dt (1/4)∫((t^2 +2t+5)/(t(t+1)))dt deviding t^2 +2t+5 by t^2 +t (1/4)∫{1+((t+5)/(t^2 +t))}dt (1/4)∫tdt+(1/8)∫((2t+1+9)/(t^2 +t))dt (1/4)∫dt+(1/8)∫((2t+1)/(t^2 +t))dt+(9/8)∫(dt/(t(t+1))) (1/4)∫dt+(1/8)∫((2t+1)/(t^2 +1))dt+(9/8)∫((t+1−t)/(t(t+1)))dt (1/4)∫dt+(1/8)∫((2t+1)/(t^2 +t))dt+(9/8)∫(dt/t)−(9/8)∫(dt/(t+1)) (1/4)t+(1/8)ln(t^2 +t)+(9/8)lnt−(9/8)ln(t+1) (1/4)(2e^x −1)+(1/8)ln{(2e^x −1)^2 +(2e^x −1}− (9/8)ln(2e^x −1+1) (1/4)(2e^x −1)+(1/8)ln(4e^(2x) −4e^x +1+2e^x −1)− (9/8){ln2+x} (1/4)(2^ e^x −1)+(1/8)ln(4e^(2x) −2e^x )+((9x)/8)+constant (1/4)(2e^x −1)+(1/8){ln2e^x (2e^x −1)}+((9x)/8)+C =(1/4)(2e^x −1)+(1/8){ln2+x+ln(2e^x −1)}+((9x)/8)+C =(1/4)(2e^x −1)+((ln2)/8)+(x/8)+(1/8)ln(2e^x −1)+((9x)/8)+C =(1/4)(2e^x −1)+((10x)/8)+(1/8)ln(2e^x −1)+C′ ans$
	$t = 2 e^{x} - 1$ $e^{x} = \frac{t + 1}{2} s o e^{x} d x = \frac{1}{2} d t$ $d x = \frac{1}{2} \times \frac{2}{t + 1} \times d t$ $\int \frac{{(\frac{t + 1}{2})}^{2} + 1}{t} \times \frac{d t}{t + 1}$ $\int \frac{t^{2} + 2 t + 1 + 4}{4 t (t + 1)} d t$ $\frac{1}{4} \int \frac{t^{2} + 2 t + 5}{t (t + 1)} d t$ $d e v i d i n g t^{2} + 2 t + 5 b y t^{2} + t$ $\frac{1}{4} \int {1 + \frac{t + 5}{t^{2} + t}} d t$ $\frac{1}{4} \int t d t + \frac{1}{8} \int \frac{2 t + 1 + 9}{t^{2} + t} d t$ $\frac{1}{4} \int d t + \frac{1}{8} \int \frac{2 t + 1}{t^{2} + t} d t + \frac{9}{8} \int \frac{d t}{t (t + 1)}$ $\frac{1}{4} \int d t + \frac{1}{8} \int \frac{2 t + 1}{t^{2} + 1} d t + \frac{9}{8} \int \frac{t + 1 - t}{t (t + 1)} d t$ $\frac{1}{4} \int d t + \frac{1}{8} \int \frac{2 t + 1}{t^{2} + t} d t + \frac{9}{8} \int \frac{d t}{t} - \frac{9}{8} \int \frac{d t}{t + 1}$ $\frac{1}{4} t + \frac{1}{8} l n (t^{2} + t) + \frac{9}{8} l n t - \frac{9}{8} l n (t + 1)$ $\frac{1}{4} (2 e^{x} - 1) + \frac{1}{8} l n {{(2 e^{x} - 1)}^{2} + (2 e^{x} - 1} -$ $\frac{9}{8} l n (2 e^{x} - 1 + 1)$ $\frac{1}{4} (2 e^{x} - 1) + \frac{1}{8} l n (4 e^{2 x} - 4 e^{x} + 1 + 2 e^{x} - 1) -$ $\frac{9}{8} {l n 2 + x}$ $\frac{1}{4} (2^{} e^{x} - 1) + \frac{1}{8} l n (4 e^{2 x} - 2 e^{x}) + \frac{9 x}{8} + c o n s t a n t$ $\frac{1}{4} (2 e^{x} - 1) + \frac{1}{8} {l n 2 e^{x} (2 e^{x} - 1)} + \frac{9 x}{8} + C$ $= \frac{1}{4} (2 e^{x} - 1) + \frac{1}{8} {l n 2 + x + l n (2 e^{x} - 1)} + \frac{9 x}{8} + C$ $= \frac{1}{4} (2 e^{x} - 1) + \frac{l n 2}{8} + \frac{x}{8} + \frac{1}{8} l n (2 e^{x} - 1) + \frac{9 x}{8} + C$ $= \frac{1}{4} (2 e^{x} - 1) + \frac{10 x}{8} + \frac{1}{8} l n (2 e^{x} - 1) + C^{'} a n s$
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