Indefinite integral $\int \ln(1-x)dx$

Question

When asked to find the indefinite integral $$\int \ln(1-x)dx$$ my first instinct is to use a u-substitution and let $u=1-x$ and therefore $-du=dx$ And after the substitution we have the indefinite integral with u as the variable of integration $-\int \ln(u)du$ which is, using integration by parts, $-u(\ln(u)-1)$ and now back-substituting for $u=1-x$ we have $-(1-x)(\ln(1-x)-1)$ which can be written as $(x-1)(\ln(1-x)-1)+c$ remembering the constant of integration. My question is, is this on the right track? Is there a more efficient or more zazzy way to go about this? Thank you in advance.

Answer 1

Taking $u=1-x\implies\mathrm du=-\mathrm dx$

$$\begin{align}\int\ln(1-x)\,\mathrm dx&=-\int\ln u\,\mathrm du\\&=-(u\ln u-u)+C\qquad\text{Integrate by parts}\\&=-u\ln u+u+C\\&=-(1-x)\ln(1-x)+(1-x)+C\\&=-(1-x)\left\{\ln(1-x)-1\right\}+C\\&=(x-1)\left\{\ln(1-x)-1\right\}+C\end{align}$$

I did the entire problem, I don't see a fancy way of doing it, just regular substitution followed by integration by parts.

Answer 2

Substituting $u=1-x$, $\frac{\mathrm{d}u}{\mathrm{d}x}=-1$, $\mathrm{d}u=-\mathrm{d}x$ $$\int{\ln{(1-x)}}\mathrm{d}x$$ $$=-\int{\ln{(u)}}\mathrm{d}u$$ By IGB: $$f(u)=\ln{(u)},f’(u)=\frac{1}{u}$$ $$g(u)=u,g’(u)=1$$ Therefore, the integral becomes: $$-u\ln{(u)}+\int{\frac{u}{u}}\mathrm{d}u$$ $$=-u\ln{(u)}+u+C$$ $$=(x-1)(\ln{(1-x)}-1)+C$$, where C is the constant of integration

I don’t think there is any faster way, as this is a simple enough integral.