# How do you evaluate the integral int ln(x^2sqrt(4x-1))?

Dec 28, 2016

By the laws of logarithms:

$= \int \ln {x}^{2} + \ln \sqrt{4 x - 1} \mathrm{dx}$

$= \int \ln {x}^{2} \mathrm{dx} + \int \ln \sqrt{4 x - 1} \mathrm{dx}$

$= 2 \int \ln x \mathrm{dx} + \frac{1}{2} \int \ln \left(4 x - 1\right) \mathrm{dx}$

We let $u = 4 x - 1$. Then $\mathrm{du} = 4 \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{4}$.

$= 2 \int \ln x \mathrm{dx} + \frac{1}{2} \int \ln u \frac{\mathrm{du}}{4}$

$= 2 \int \ln x \mathrm{dx} + \frac{1}{8} \int \ln u \mathrm{du}$

We need to integrate the natural logarithm function by parts. Let $u = \ln x$ and $\mathrm{dv} = 1 \mathrm{dx}$. Then $\mathrm{du} = \frac{1}{x} \mathrm{dx}$ and $v = x$.

$\int \left(u \mathrm{dv}\right) = u v - \int \left(v \mathrm{du}\right)$

$\int \left(\ln x\right) = x \ln x - \int \left(x \cdot \frac{1}{x}\right) \mathrm{dx}$

$\int \left(\ln x\right) = x \ln x - x + C$

Back to the original integral.

$= 2 \left(x \ln x - x\right) + \frac{1}{8} \left(u \ln u - u\right) + C$

$= 2 x \ln x - 2 x + \frac{1}{8} \left(4 x - 1 \ln \left(4 x - 1\right) - \left(4 x - 1\right)\right) + C$

$= 2 x \ln x - 2 x + \frac{\left(4 x - 1\right) \ln \left(4 x - 1\right) - 4 x + 1}{8} + C$

Hopefully this helps!