calculate ∫_(−1) ^1 ((ln(x+2))/((x+4)^2 −1))dx


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Question Number 47248 by maxmathsup by imad last updated on 06/Nov/18

$c a l c u l a t e \int_{- 1}^{1} \frac{l n (x + 2)}{{(x + 4)}^{2} - 1} d x$
Commented by maxmathsup by imad last updated on 16/Nov/18
$let A = ∫_(−1) ^1 ((ln(x+2))/((x+4)^2 −1))dx ⇒ A =_(x+2=t) ∫_1 ^3 ((ln(t))/((t+2)^2 −1))dt =∫_1 ^3 ((ln(t))/((t+1)(t+3)))dt =(1/2)∫_1 ^3 {(1/(t+1)) −(1/(t+3))}ln(t)dt =(1/2) ∫_1 ^3 ((ln(t))/(t+1))dt −(1/2) ∫_1 ^3 ((ln(t))/(t+3))dt let determine ∫ ((ln(x))/(x+1))dx the calculator give ∫ ((ln(x))/(x+1)) dx=ln(x)ln(x+1) +L_(i 2) (−x) ⇒ ∫_1 ^3 ((ln(t))/(t+1))dt =[ln(t)ln(t+1)+L_(i2) (−t)]_1 ^3 =2ln(3)ln(2) +L_(i2) (−3) −L_(i2) (−1) also ∫ ((ln(x))/(x+3))dx =ln(x)ln((x/3)+1)+L_(i2) (−(x/3)) ⇒ ∫_1 ^3 ((ln(x))/(x+3)) =[ln(x)ln((x/3)+1)+L_(i2) (−(x/3))]_1 ^3 =ln(3)ln(2)+L_(i2) (−1)−L_(i2) (−(1/3))....$
$l e t A = \int_{- 1}^{1} \frac{l n (x + 2)}{{(x + 4)}^{2} - 1} d x \Rightarrow A =_{x + 2 = t} \int_{1}^{3} \frac{l n (t)}{{(t + 2)}^{2} - 1} d t$ $= \int_{1}^{3} \frac{l n (t)}{(t + 1) (t + 3)} d t = \frac{1}{2} \int_{1}^{3} {\frac{1}{t + 1} - \frac{1}{t + 3}} l n (t) d t$ $= \frac{1}{2} \int_{1}^{3} \frac{l n (t)}{t + 1} d t - \frac{1}{2} \int_{1}^{3} \frac{l n (t)}{t + 3} d t l e t d e t e r m i n e \int \frac{l n (x)}{x + 1} d x t h e c a l c u l a t o r$ $g i v e \int \frac{l n (x)}{x + 1} d x = l n (x) l n (x + 1) + L_{i 2} (- x) \Rightarrow$ $\int_{1}^{3} \frac{l n (t)}{t + 1} d t = {[l n (t) l n (t + 1) + L_{i 2} (- t)]}_{1}^{3} = 2 l n (3) l n (2) + L_{i 2} (- 3)$ $- L_{i 2} (- 1) a l s o \int \frac{l n (x)}{x + 3} d x = l n (x) l n (\frac{x}{3} + 1) + L_{i 2} (- \frac{x}{3}) \Rightarrow$ $\int_{1}^{3} \frac{l n (x)}{x + 3} = {[l n (x) l n (\frac{x}{3} + 1) + L_{i 2} (- \frac{x}{3})]}_{1}^{3}$ $= l n (3) l n (2) + L_{i 2} (- 1) - L_{i 2} (- \frac{1}{3}) . . . .$
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