## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{3}{2}\ln(x^{2}+1)-3\tan^{-1}x+\sqrt{2}\tan^{-1}(\frac{x}{\sqrt{2}})+C$
The degree of the numerator is 3, the denominator's is 4. Try partial fractions. The denominator has two prime quadratic factors. $\displaystyle \frac{3x^{3}-x^{2}+6x-4}{(x^{2}+1)(x^{2}+2)}=\frac{Ax+B}{(x^{2}+1) }+\frac{Cx+D}{(x^{2}+2)}$ $3x^{3}-x^{2}+6x-4=(Ax+B)(x^{2}+2)+(Cx+D)(x^{2}+1)$ $3x^{3}-x^{2}+6x-4=(A+C)x^{3}+(B+D)x^{2}+(2A+C)x+(2B+D)$ Equating coefficients of $x^{3}$ and $x: \left\{\begin{array}{l} A+C=3\\ 2A+C=6 \end{array}\right.$ Subtract the first equation from the second: $A=3, C=0$ Equating coefficients of $x^{2}$ and the linear term$:$ $\left\{\begin{array}{l} B+D=-1\\ 2B+D=-4 \end{array}\right.\Rightarrow B=-3,D=2$ $I=\displaystyle \int\frac{3x^{3}-x^{2}+6x-4}{(x^{2}+1)(x^{2}+2)}dx= \int\frac{3x-3}{x^{2}+1 }dx+\int\frac{2}{x^{2}+2}dx$ $= \displaystyle \frac{3}{2}\int\frac{2x }{x^{2}+1 }dx-3\int\frac{dx}{x^{2}+1 }+2\int\frac{dx}{x^{2}+2}$ ... first integral : $\displaystyle \left[\begin{array}{l} t=x^{2}+1\\ dt=2xdx \end{array}\right]\Rightarrow I_{1}\int\frac{dt}{t}=\ln|t|+C$ ... for the second and third, table integral 17: $\displaystyle \int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\tan^{-1}\frac{u}{a}+C$ $I==\displaystyle \frac{3}{2}\ln(x^{2}+1)-3\tan^{-1}x+2\cdot\frac{1}{\sqrt{2}}\tan^{-1}(\frac{x}{\sqrt{2}})+C$ $=\displaystyle \frac{3}{2}\ln(x^{2}+1)-3\tan^{-1}x+\sqrt{2}\tan^{-1}(\frac{x}{\sqrt{2}})+C$