Integrals containing the form $ a + bu$ .

$\displaystyle \int \frac {u \; du}{a + bu} = \frac {1}{b^2} (a + bu - a \ln \vert a + bu \vert ) + C.$ (B..1)

$\displaystyle \int \frac{u^2 \; du}{a + bu} = \frac {1}{b^3} \left[ \frac12 (a + bu)^2 - 2a(a + bu) + a^2 \ln \vert a + bu \vert \right]+ C.$ (B..2)

$\displaystyle \int \frac{u \; du}{(a + bu)^2} = \frac {1}{b^2} \left( \frac{a}{a + bu} + \ln \vert a + bu \vert \right)+ C.$ (B..3)

$\displaystyle \int \frac{u^2 \;du}{(a + bu)^2} = \frac {1}{b^3} \left( a + bu - \frac {a^2}{a + bu} - 2a \ln \vert a + bu \vert\right) + C.$ (B..4)

$\displaystyle \int \frac {u \; du}{(a + bu)^3} = \frac {1}{b^2} \left( \frac{a}{2(a + bu)^2} - \frac {1}{a + bu} \right) +C.$ (B..5)

$\displaystyle \int \frac {du}{u(a + bu)} = \frac 1a \ln \begin{vmatrix}\frac {u}{a + bu} \end{vmatrix}+C.$ (B..6)

$\displaystyle \int \frac {du}{u^2 (a + bu)} = -\frac {1}{au} + \frac{b}{a^2} \ln \begin{vmatrix}\frac{a + bu}{u} \end{vmatrix}+ C.$ (B..7)

$\displaystyle \int \frac {du}{u(a + bu)^2} = \frac{1}{a(a + bu)} + \frac{1}{a^2} \ln \begin{vmatrix}\frac{u}{a + bu} \end{vmatrix} + C.$ (B..8)

Noah Dana-Picard 2007-12-28