Integral of rational function with a squared term in the denominator

Question

I know the integration when in the reciprocal there's only degree $1$, but what about degree $2$? Take an example, $$\int\frac{x \, \mathrm{d}x}{a+bx^2}$$

Answer 1

For this, put $v=a+bx^{2}$ then $dv = 2bxdx$ from which \begin{eqnarray} \int \frac{xdx}{a+bx^{2}} &=& \frac{1}{2b}\int \frac{1}{v}dv \\ &=& \frac{1}{2b}\ln | v |+ c \\ &=& \frac{1}{2b}\ln |a+bx^{2}|+c \\ \end{eqnarray}

Answer 2

$$\int\frac{xdx}{a+bx^2}=\frac1{2b}\int\frac{2bxdx}{a+bx^2}=\frac1{2b}\int\frac{d(a+bx^2)}{a+bx^2}=\frac1{2b}\ln(a+bx^2)+C$$

Answer 3

Let $u = a+bx^2$ so that $\mathrm{d}u =2bx \, \mathrm{d}x \implies {\color{blue}{x \, \mathrm{d}x = \frac{1}{2b} \, \mathrm{d}u}}$. Then our integral becomes $$\int \frac{\color{blue}{x \, \mathrm{d}x}}{a+bx^2} = \frac{1}{2b}\int \frac{1}{u} \, \mathrm{d}u$$

This evaluates to $$\frac{1}{2b} \ln |u| + \mathrm{C} = \frac{1}{2b} \ln |a+bx^2| + \mathrm{C}$$