int frac{dx}{x(x+1)} equals,where c is an arb | vedclass

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(B) To evaluate the integral $\int \frac{dx}{x(x+1)}$,we use the method of partial fractions.We can write the integrand as: $\frac{1}{x(x+1)} = \frac{

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$\ln \left|\frac{x}{x+1}\right|+c$

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(B) To evaluate the integral $\int \frac{dx}{x(x+1)}$,we use the method of partial fractions.We can write the integrand as: $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$.Multiplying by $x(x+1)$,we get $1 = A(x+1) + Bx$.Setting $x = 0$,we find $A = 1$.Setting $x = -1$,we find $B = -1$.Thus,$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$.Now,integrate each term: $\int \left(\frac{1}{x} - \frac{1}{x+1}\right) dx = \int \frac{1}{x} dx - \int \frac{1}{x+1} dx$.This results in $\ln |x| - \ln |x+1| + c$.Using the logarithmic property $\ln a - \ln b = \ln \left|\frac{a}{b}\right|$,we get $\ln \left|\frac{x}{x+1}\right| + c$.

$\int \frac{dx}{x(x+1)}$ equals,where $c$ is an arbitrary constant.

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