Why Does My Integral Calculation Yield the Inverse Result?

[jiasyuen]

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

$$=\int \frac{-4}{x}dx-\int \frac{1}{1-x}dx$$

$$=-4\int \frac{1}{x}dx-\int\frac{1}{1-x}dx$$

$$=-4\ln\left | x \right |-\ln \left | 1-x \right |+c$$

$$\ln \frac{x^4}{\left | 1-x \right |}+c$$

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

Where's my mistake?

 

[MarkFL]

I agree with you up to here:

$$-4\int \frac{1}{x}\,dx-\int\frac{1}{1-x}\,dx$$

Then upon integrating, we obtain:

$$-4\ln|x|+\ln|1-x|+C$$

And then combining the two log terms, we get:

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

 

[SuperSonic4]

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

$$=\int \frac{-4}{x}dx-\int \frac{1}{1-x}dx$$

$$=-4\int \frac{1}{x}dx-\int\frac{1}{1-x}dx$$

$$=-4\ln\left | x \right |-\ln \left | 1-x \right |+c$$

$$\ln \frac{x^4}{\left | 1-x \right |}+c$$

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

Where's my mistake?

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

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