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

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(B) The integral is $I = \int \frac{dx}{(x + 1)^2 (x^2 + 1)}$.Using partial fractions,we write $\frac{1}{(x + 1)^2 (x^2 + 1)} = \frac{A}{(x + 1)^2} +

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$\log_e \sqrt{x + 1} - \frac{1}{2} \log_e \sqrt{x^2 + 1} - \frac{1}{2(x + 1)} + C$

Vedclass · Answer

(B) The integral is $I = \int \frac{dx}{(x + 1)^2 (x^2 + 1)}$.Using partial fractions,we write $\frac{1}{(x + 1)^2 (x^2 + 1)} = \frac{A}{(x + 1)^2} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 1}$.Multiplying by $(x + 1)^2 (x^2 + 1)$,we get $1 = A(x^2 + 1) + B(x + 1)(x^2 + 1) + (Cx + D)(x + 1)^2$.Setting $x = -1$,we find $A = \frac{1}{2}$.Comparing coefficients of $x^3$,$0 = B + C$. Comparing coefficients of $x^2$,$0 = A + B + D + 2C$. Comparing coefficients of $x$,$0 = B + 2D + C$. Constant term $1 = A + B + D$.Solving these,we get $A = \frac{1}{2}$,$B = \frac{1}{2}$,$C = -\frac{1}{2}$,$D = 0$.Thus,$I = \int [\frac{1}{2(x + 1)^2} + \frac{1}{2(x + 1)} - \frac{x}{2(x^2 + 1)}] dx$.Integrating term by term,$I = \frac{1}{2} [-\frac{1}{x + 1} + \log_e|x + 1| - \frac{1}{2} \log_e(x^2 + 1)] + C$.$I = \frac{1}{2} \log_e(x + 1) - \frac{1}{4} \log_e(x^2 + 1) - \frac{1}{2(x + 1)} + C$.Using $\frac{1}{2} \log_e(x + 1) = \log_e \sqrt{x + 1}$ and $\frac{1}{4} \log_e(x^2 + 1) = \frac{1}{2} \log_e \sqrt{x^2 + 1}$,we get $I = \log_e \sqrt{x + 1} - \frac{1}{2} \log_e \sqrt{x^2 + 1} - \frac{1}{2(x + 1)} + C$.

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

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