Evaluate $\int \frac{x^3 + 3x + 2}{(x^2+1)^2(x+1)} dx$. The result can be expressed as $\frac{1}{4}\

Question

Evaluate $\int \frac{x^3 + 3x + 2}{(x^2+1)^2(x+1)} dx$. The result can be expressed as $\frac{1}{4}\ln\left(\frac{x^2+1}{(x+1)^2}\right) + f(x) + C$. Which of the following represents $f(x)$?

Professor PC Thomas Classes · Answer

Split numerator: $\frac{x(x^2+1) + 2(x+1)}{(x^2+1)^2(x+1)} = \frac{x}{(x^2+1)(x+1)} + \frac{2}{(x^2+1)^2}$. Partial fractions for 1st term: $\frac{-1/2}{x+1} + \frac{x+1}{2(x^2+1)}$, which integrates to $-\frac{1}{2}\ln|x+1| + \frac{1}{4}\ln(x^2+1) + \frac{1}{2}	an^{-1}x$. For the 2nd term, substitute $x = 	an	heta \implies \int \frac{2\sec^2	heta}{\sec^4	heta}d	heta = 2\int\cos^2	heta d	heta = 	heta + \frac{1}{2}\sin 2	heta = 	an^{-1}x + \frac{x}{x^2+1}$. Adding them up gives the required form with $f(x) = \frac{3}{2}	an^{-1} x + \frac{x}{x^2+1}$.