# Thread: Indefinite Integral Question

1. ## Indefinite Integral Question

Hi guys,

I was asked a question about an indefinite integral and have no idea where to begin. I tried playing around with the numbers but have not had success.

The question is as follows: Evaluate the indefinite integral - Int [12ln(3x-5) / (3x-5)]dx.
I couldn't figure out how to put the integral symbol in so "Int" is what I used to represent it.

Any help would be appreciated.

Thanks.

2. ## Re: Indefinite Integral Question

Notice that if you substitute \displaystyle \begin{align*} u = \ln{ (3x - 5)} \end{align*} then \displaystyle \begin{align*} \mathrm{d}u = \frac{3}{3x - 5}\,\mathrm{d}x \end{align*}, so write your integral as

\displaystyle \begin{align*} \int{ \frac{12\ln{(3x-5)}}{3x-5}\,\mathrm{d}x} &= 4\int{ \ln{(3x-5)}\,\frac{3}{3x-5}\,\mathrm{d}x } \\ &= 4\int{ u\,\mathrm{d}u} \\ &= 4 \left( \frac{u^2}{2} \right) + C \\ &= 2u^2 + C \\ &= 2\left[ \ln{(3x-5)} \right] ^2 + C \end{align*}

3. ## Re: Indefinite Integral Question

Very clear and to the point.

Thank you very much!