Evaluating indefinite integral $\int \frac{3x}{x-2}\,dx$

Question

$$\int \frac{3x}{x-2}\,dx$$

The answer is $3(2\ln|x-2|+x)+C\,$, but I don't understand how this is the answer. I thought I could just separate the $x$ on the numerator from the equation and evaluate them separately to get $$\frac{3x^2\ln\left|x-2\right|}{2}$$ as the answer. Why doesn't my approach work?

Answer 1

Hint: try writing the numerator instead as $3x-6+6$ and break it into two nice terms.

Alternatively, try a change of variable $u = x-2$.

Answer 2

The integral of a product is not the product of the integrals: $$\begin{align} \int 3x \;dx &= \frac{3x^2}{2} + C\\ \int \frac{1}{x-2}dx &= \ln|x-2| + C\\ \int \frac{3x}{x-2} \;dx &\neq \frac{3x^2\ln|x-2|}{2} + C \end{align}$$ Try to do something like this instead: $$\begin{align} \int \frac{3x}{x-2} \;dx &= \int \left(3 + \frac{6}{x-2}\right) \;dx\\ &= \int 3\;dx + \int \frac{6}{x-2} \;dx\\ &= 3x + 6\ln|x-2| + C \end{align}$$

Answer 3

I would like to propose another method, for the education of the OP. By using long division and doing the following operation, you end up with a simplified answer.

$$\require{enclose}x\enclose{longdiv}{x-2}$$

You end up with $1+\dfrac{2}{x-2}$ which is very easy to integrate.

Answer 4

Write it as

$$3\int\frac{x}{x-2}dx = 3\int\frac{x-2+2}{x-2}dx = 3\int 1 + \frac{2}{x-2} dx =3[ x + 2\ln(x + 2)] + C =\dots$$

This gives the final answer

Answer 5

Or alternatively, try a $u$-substitution. Let $u = x-2$, then $dx = du$ and $x = u + 2$, so your integral becomes

$$3\int \dfrac{u+2}{u} du = 3\int \left(1 + \frac{2}{u}\right)du =\quad ...$$

Answer 6

here is my answer

$\int\frac{3x}{x-2}dx$

$=\int\frac{3(x-2)+6}{x-2}dx$

$=3\int dx+6\int\frac{1}{x-2}dx$

$=3x+6\ln(x-2)+c$

c is cosntant of integration