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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int \frac{x^2 + 1}{x(x^2 - 1)} \, dx$.Divide the numerator and denominator by $x^2$:$I = \int \frac{1 + \frac{1}{x^2}}{x - \frac{1}{x}}

Vedclass · Accepted Answer

$\log \left| \frac{x^2 - 1}{x} \right| + c$

Vedclass · Answer

(A) Let $I = \int \frac{x^2 + 1}{x(x^2 - 1)} \, dx$.Divide the numerator and denominator by $x^2$:$I = \int \frac{1 + \frac{1}{x^2}}{x - \frac{1}{x}} \, dx$.Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) \, dx$.Substituting these into the integral,we get:$I = \int \frac{1}{t} \, dt = \log |t| + c$.Substituting back $t = x - \frac{1}{x} = \frac{x^2 - 1}{x}$,we get:$I = \log \left| \frac{x^2 - 1}{x} \right| + c$.

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

Similar Questions

Integrate the function: $\frac{x^{2}}{1-x^{6}}$

$\int \frac{dx}{1 + 3\sin^2 x} = $

The value of $\int {\frac{{\sqrt {{x^2} - {a^2}} }}{x}dx} $ is:

$\int \frac{\cos 2x + x + 1}{x^2 + \sin 2x + 2x} \, dx = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module