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

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Question

(C) Let $I = \int \frac{(x+1)^2}{x(x^2+1)} dx$. Expanding the numerator,we get $(x+1)^2 = x^2 + 2x + 1$. So,$I = \int \frac{x^2 + 2x + 1}{x(x^2+1)} dx

Vedclass · Accepted Answer

$\log |x| + 2 	an^{-1}(x) + C$

Vedclass · Answer

(C) Let $I = \int \frac{(x+1)^2}{x(x^2+1)} dx$. Expanding the numerator,we get $(x+1)^2 = x^2 + 2x + 1$. So,$I = \int \frac{x^2 + 2x + 1}{x(x^2+1)} dx$. We can rewrite the numerator as $(x^2+1) + 2x$. Thus,$I = \int \frac{(x^2+1) + 2x}{x(x^2+1)} dx$. Splitting the integral,we get $I = \int \frac{x^2+1}{x(x^2+1)} dx + \int \frac{2x}{x(x^2+1)} dx$. $I = \int \frac{1}{x} dx + 2 \int \frac{1}{x^2+1} dx$. Integrating,we get $I = \log |x| + 2 	an^{-1}(x) + C$. Hence,option $(C)$ is correct.

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

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