int frac{(x + 1)^2}{x(x^2 + 1)} , dx is equal | vedclass

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(B) We are given the integral $I = \int \frac{(x + 1)^2}{x(x^2 + 1)} \, dx$. Expanding the numerator,we get $(x + 1)^2 = x^2 + 1 + 2x$. Substituting t

Vedclass · Accepted Answer

$\log_e x + 2	an^{-1} x + c$

Vedclass · Answer

(B) We are given the integral $I = \int \frac{(x + 1)^2}{x(x^2 + 1)} \, dx$. Expanding the numerator,we get $(x + 1)^2 = x^2 + 1 + 2x$. Substituting this into the integral,we have $I = \int \frac{x^2 + 1 + 2x}{x(x^2 + 1)} \, dx$. Splitting the integral,we get $I = \int \frac{x^2 + 1}{x(x^2 + 1)} \, dx + \int \frac{2x}{x(x^2 + 1)} \, dx$. Simplifying the terms,we get $I = \int \frac{1}{x} \, dx + 2 \int \frac{1}{x^2 + 1} \, dx$. Integrating these standard forms,we obtain $I = \log_e |x| + 2	an^{-1} x + c$.

$\int \frac{(x + 1)^2}{x(x^2 + 1)} \, dx$ is equal to

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