9cii please!
$\displaystyle \int ln(4x-3)dx$
$\displaystyle \int ln(4x-3).dx$
. means multiply okok
u= ln(4x-3) du/dx= 4/4x-3
dV/dx= 1, V= x
sooooo
$\displaystyle \int ln(4x-3)dx = xln(4x-3) - \int x.\frac{4}{4x-3}$
$\displaystyle \int ln(4x-3)dx = xln(4x-3) - \int \frac{4x}{4x-3}$
second integral look familiar ? Its the question before with substitution how lovely!
$\displaystyle \int ln(4x-3)dx = xln(4x-3) - \frac{4x-2}{4}+ \frac{3}{4}ln(4x-3) + C$