int frac{x - 1}{(x + 1)^2} , dx = | vedclass

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(A) To evaluate the integral $I = \int \frac{x - 1}{(x + 1)^2} \, dx$,we rewrite the numerator in terms of the denominator: $x - 1 = (x + 1) - 2$. Sub

Vedclass · Accepted Answer

$\log |x + 1| + \frac{2}{x + 1} + C$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{x - 1}{(x + 1)^2} \, dx$,we rewrite the numerator in terms of the denominator: $x - 1 = (x + 1) - 2$. Substituting this into the integral: $I = \int \frac{(x + 1) - 2}{(x + 1)^2} \, dx$ $I = \int \left( \frac{x + 1}{(x + 1)^2} - \frac{2}{(x + 1)^2} \right) \, dx$ $I = \int \frac{1}{x + 1} \, dx - 2 \int (x + 1)^{-2} \, dx$ Using the standard integration formulas $\int \frac{1}{u} \, du = \log |u| + C$ and $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$: $I = \log |x + 1| - 2 \left( \frac{(x + 1)^{-1}}{-1} \right) + C$ $I = \log |x + 1| + \frac{2}{x + 1} + C$.

$\int \frac{x - 1}{(x + 1)^2} \, dx = $

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