calculate I = ∫_0 ^1 e^(2t) ln(1+e^t )dt


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Question Number 35684 by prof Abdo imad last updated on 22/May/18

$c a l c u l a t e I = \int_{0}^{1} e^{2 t} l n (1 + e^{t}) d t$
Commented by prof Abdo imad last updated on 22/May/18
$let put I = ∫_0 ^1 e^(2t) ln(1+e^t )dt .changement e^t =x give I = ∫_1 ^e x^2 ln(1+x)(dx/x) = ∫_1 ^e xln(1+x)dx .by parts we get I = [ (x^2 /2)ln(1+x)]_1 ^e −∫_1 ^e (x^2 /2) (dx/(1+x)) =(e^2 /2)ln(1+e) −(1/2)ln(2) −(1/2) ∫_1 ^e (x^2 /(1+x))dx ∫_1 ^e (x^2 /(x+1))dx = ∫_1 ^e ((x^2 −1 +1)/(x+1))dx = ∫_1 ^e (x−1)dx + ∫_1 ^e (dx/(x+1)) =[ (x^2 /2) −x]_1 ^e + [ln(x+1)]_1 ^e =(e^2 /2) −e −(1/2) +1 +ln(1+e) −ln(2) ⇒ I = (e^2 /2)ln(1+e) −(1/2)ln(2)−(1/4)e^2 +(1/2)e −(1/4) +ln(1+e) −ln(2) ={ 1+(e^2 /2)}ln(1+e) −(3/2)ln(2) −(e^2 /4) +(e/2)−(1/4).$
$l e t p u t I = \int_{0}^{1} e^{2 t} l n (1 + e^{t}) d t . c h a n g e m e n t$ $e^{t} = x g i v e I = \int_{1}^{e} x^{2} l n (1 + x) \frac{d x}{x}$ $= \int_{1}^{e} x l n (1 + x) d x . b y p a r t s w e g e t$ $I = {[\frac{x^{2}}{2} l n (1 + x)]}_{1}^{e} - \int_{1}^{e} \frac{x^{2}}{2} \frac{d x}{1 + x}$ $= \frac{e^{2}}{2} l n (1 + e) - \frac{1}{2} l n (2) - \frac{1}{2} \int_{1}^{e} \frac{x^{2}}{1 + x} d x$ $\int_{1}^{e} \frac{x^{2}}{x + 1} d x = \int_{1}^{e} \frac{x^{2} - 1 + 1}{x + 1} d x$ $= \int_{1}^{e} (x - 1) d x + \int_{1}^{e} \frac{d x}{x + 1}$ $= {[\frac{x^{2}}{2} - x]}_{1}^{e} + {[l n (x + 1)]}_{1}^{e}$ $= \frac{e^{2}}{2} - e - \frac{1}{2} + 1 + l n (1 + e) - l n (2) \Rightarrow$ $I = \frac{e^{2}}{2} l n (1 + e) - \frac{1}{2} l n (2) - \frac{1}{4} e^{2} + \frac{1}{2} e - \frac{1}{4}$ $+ l n (1 + e) - l n (2)$ $= {1 + \frac{e^{2}}{2}} l n (1 + e) - \frac{3}{2} l n (2) - \frac{e^{2}}{4} + \frac{e}{2} - \frac{1}{4} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18
	$=∫_0 ^1 e^(2t) (e^t −(e^(2t) /2)+(e^(3t) /3)−((e^(4t) )/4)+...)dt =∫_0 ^1 e^(3t) −(e^(4t) /2)+(e^(5t) /3)−(e^(6t_ ) /4)+... dt =∣(e^(3t) /3)−(e^(4t) /(2×4))+(e^(5t) /(3×5))−(e^(6t) /(4×6))+....∣_0 ^1 =((e^3 /3)−(e^4 /(2×4))+(e^5 /(3×5))−(e^6 /(4×6))...)−( (1/3)−(1/(2×4))+(1/(3×5)) −(1/(4×6))+....) let s_1 =sum total of first s_(2 ) =sum total of second T_n =(−1)^(n−1) {(1/(n×(n+2)))} general term for s_2 T_n =(1/2)×(−1)^(n−1) ×{(1/n)−(1/(n+2))} T_1 =(1/2)×1×{(1/1)−(1/3)} T_2 =(1/2)×1×{(1/4)−(1/2)} T_3 =(1/2)×1×{(1/3)−(1/5)} .......... ........... adding we get (1/2)×1×{1−(1/2)} other terms cancelled so s_2 =(1/2)×(1/2)=(1/4) contd to find s_1 T_n =(−1)^(n−1) ×{(e^(n+2) /(n×(n+2)))} =(−1)^(n−1) ×e^(n+2) ×(1/2)×((1/n)−(1/(n+2))) contd$
	$= \int_{0}^{1} e^{2 t} (e^{t} - \frac{e^{2 t}}{2} + \frac{e^{3 t}}{3} - \frac{e^{4 t}}{4} + . . .) d t$ $= \int_{0}^{1} e^{3 t} - \frac{e^{4 t}}{2} + \frac{e^{5 t}}{3} - \frac{e^{6 t_{}}}{4} + . . . d t$ $=∣ \frac{e^{3 t}}{3} - \frac{e^{4 t}}{2 \times 4} + \frac{e^{5 t}}{3 \times 5} - \frac{e^{6 t}}{4 \times 6} + . . . . ∣_{0}^{1}$ $= (\frac{e^{3}}{3} - \frac{e^{4}}{2 \times 4} + \frac{e^{5}}{3 \times 5} - \frac{e^{6}}{4 \times 6} . . .) - (\frac{1}{3} - \frac{1}{2 \times 4} + \frac{1}{3 \times 5}$ $- \frac{1}{4 \times 6} + . . . .)$ $l e t s_{1} = s u m t o t a l o f f i r s t$ $s_{2} = s u m t o t a l o f s e c o n d$ $T_{n} = {(- 1)}^{n - 1} {\frac{1}{n \times (n + 2)}} g e n e r a l t e r m f o r s_{2}$ $T_{n} = \frac{1}{2} \times {(- 1)}^{n - 1} \times {\frac{1}{n} - \frac{1}{n + 2}}$ $T_{1} = \frac{1}{2} \times 1 \times {\frac{1}{1} - \frac{1}{3}}$ $T_{2} = \frac{1}{2} \times 1 \times {\frac{1}{4} - \frac{1}{2}}$ $T_{3} = \frac{1}{2} \times 1 \times {\frac{1}{3} - \frac{1}{5}}$ $. . . . . . . . . .$ $. . . . . . . . . . .$ $a d d i n g w e g e t \frac{1}{2} \times 1 \times {1 - \frac{1}{2}} o t h e r t e r m s$ $c a n c e l l e d$ $s o s_{2} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ $c o n t d t o f i n d s_{1}$ $T_{n} = {(- 1)}^{n - 1} \times {\frac{e^{n + 2}}{n \times (n + 2)}}$ $= {(- 1)}^{n - 1} \times e^{n + 2} \times \frac{1}{2} \times (\frac{1}{n} - \frac{1}{n + 2})$ $c o n t d$
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