1. ## Integration Query

2. Originally Posted by Kevlar

$\displaystyle \int u \, dv = uv - \int v \, du$.

Let $\displaystyle u = \ln (4x - 3)$ and $\displaystyle dv = dx$.

3. I've gotten upto

Integral ln4x-3 dx = xln4x-3 - integral x X 4/4x-3

then i get stuck :\

4. $\displaystyle \int{\ln (4x-3)\,dx}=\int{(x)'\ln (4x-3)\,dx}=x\ln (4x-3)-4\int{\frac{x}{4x-3}\,dx}.$

See my signature for further details about integration by parts.

5. Originally Posted by Kevlar
I've gotten upto

Integral ln4x-3 dx = xln4x-3 - integral x X 4/4x-3

then i get stuck :\
Note that $\displaystyle \frac{4x}{4x - 3} = \frac{(4x - 3) + 3}{4x-3} = 1 + \frac{1}{4x - 3}$.

6. $\displaystyle \int ln(4x-3)dx$

$\displaystyle \int ln(4x-3).dx$
. 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!
$\displaystyle \int ln(4x-3)dx = xln(4x-3) - \frac{4x-2}{4}+ \frac{3}{4}ln(4x-3) + C$