int_0^a frac{x-a}{x+a} dx = | vedclass

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(A) Let $I = \int_0^a \frac{x-a}{x+a} dx$. Substitute $t = x + a$,so $x = t - a$ and $dx = dt$. When $x = 0$,$t = a$. When $x = a$,$t = 2a$. Thus,$I =

Vedclass · Accepted Answer

$a - 2a \log 2$

Vedclass · Answer

(A) Let $I = \int_0^a \frac{x-a}{x+a} dx$. Substitute $t = x + a$,so $x = t - a$ and $dx = dt$. When $x = 0$,$t = a$. When $x = a$,$t = 2a$. Thus,$I = \int_a^{2a} \frac{(t-a)-a}{t} dt = \int_a^{2a} \frac{t-2a}{t} dt$. $I = \int_a^{2a} (1 - \frac{2a}{t}) dt$. $I = [t]_a^{2a} - 2a [\log |t|]_a^{2a}$. $I = (2a - a) - 2a (\log 2a - \log a)$. $I = a - 2a \log(\frac{2a}{a})$. $I = a - 2a \log 2$.

$\int_0^a \frac{x-a}{x+a} dx =$

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