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

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(B) Let $I = \int \frac{x^2+1}{(x+1)^2} dx$. We can rewrite the numerator as $x^2+1 = (x^2+2x+1) - 2x = (x+1)^2 - 2x$. So,$I = \int \frac{(x+1)^2 - 2x

Vedclass · Accepted Answer

$x-2 \log |x+1|-\frac{2}{x+1}+c$,where $c$ is a constant of integration.

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(B) Let $I = \int \frac{x^2+1}{(x+1)^2} dx$. We can rewrite the numerator as $x^2+1 = (x^2+2x+1) - 2x = (x+1)^2 - 2x$. So,$I = \int \frac{(x+1)^2 - 2x}{(x+1)^2} dx = \int 1 dx - \int \frac{2x}{(x+1)^2} dx$. For the second integral,write $2x = 2(x+1) - 2$. $I = x - \int \frac{2(x+1)-2}{(x+1)^2} dx = x - 2 \int \frac{1}{x+1} dx + 2 \int \frac{1}{(x+1)^2} dx$. Integrating these terms,we get $I = x - 2 \log |x+1| + 2 \left( -\frac{1}{x+1} \right) + c$. $I = x - 2 \log |x+1| - \frac{2}{x+1} + c$.

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

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