Why Does My Integral Calculation Yield the Inverse Result?

[jiasyuen]

\(\displaystyle \int \frac{3x-4}{x(1-x)}dx\)

\(\displaystyle =\int \frac{-4}{x}dx-\int \frac{1}{1-x}dx\)

\(\displaystyle =-4\int \frac{1}{x}dx-\int\frac{1}{1-x}dx\)

\(\displaystyle =-4\ln\left | x \right |-\ln \left | 1-x \right |+c\)

\(\displaystyle \ln \frac{x^4}{\left | 1-x \right |}+c\)

But the correct answer is \(\displaystyle \ln \frac{\left | 1-x \right |}{x^4}+c\).

Where's my mistake?

 

[MarkFL]

I agree with you up to here:

\(\displaystyle -4\int \frac{1}{x}\,dx-\int\frac{1}{1-x}\,dx\)

Then upon integrating, we obtain:

\(\displaystyle -4\ln|x|+\ln|1-x|+C\)

And then combining the two log terms, we get:

\(\displaystyle \ln\left(\frac{|1-x|}{x^4}\right)+C\)

 

[SuperSonic4]

\(\displaystyle \int \frac{3x-4}{x(1-x)}dx\)

\(\displaystyle =\int \frac{-4}{x}dx-\int \frac{1}{1-x}dx\)

\(\displaystyle =-4\int \frac{1}{x}dx-\int\frac{1}{1-x}dx\)

\(\displaystyle =-4\ln\left | x \right |-\ln \left | 1-x \right |+c\)

\(\displaystyle \ln \frac{x^4}{\left | 1-x \right |}+c\)

But the correct answer is \(\displaystyle \ln \frac{\left | 1-x \right |}{x^4}+c\).

Where's my mistake?

You have a sign error when you integrated \(\displaystyle \displaystyle - \int \frac{1}{1-x}\, dx\).

\(\displaystyle \displaystyle \int \dfrac{dx}{ax+b} = \dfrac{1}{a} \ln |ax+b|\) - in your case because \(\displaystyle a=-1\) you end up with a minus sign outside the integral