Indefinite integral. Help

Question

$\int\frac{2x+3}{x^2-4}dx=\int\frac{2x+3}{(x+2)(x-2)}dx$

So...

$A=\frac{1}{4}$

$B=\frac{7}{4}$

$\frac{1}{4}\log_{10}|x+2|+\frac{7}{4}\log_{10}|x-2|+c$

The result should be: $\log_{10}|x^2-4|+\frac{3}{4}\log_{10}|\frac{x-2}{x+2}|+c$

What did I do wrong?

Answer 1

Your computation is correct (apart from using $\log_{10}$ instead of $\log$, with implied base $e$, or $\ln$).

Probably the book chose a different approach: $$ \frac{2x+3}{x^2-4}=\frac{2x}{x^2-4}+\frac{3}{x^2-4}= \frac{2x}{x^2-4}+\frac{3}{4}\frac{1}{x-2}-\frac{3}{4}\frac{1}{x+2} $$ so integration gives $$ \log\lvert x^2-4\rvert+\frac{3}{4}\log\left|\frac{x-2}{x+2}\right|+c $$ However this is in no way “better” than your form for the antiderivative.

Answer 2

You didn't go wrong. You simply didn't rewrite the result for simplification. (B.t.w.the solution is in terms of the natural, not decimal log). Here's how it goes: $$\frac14\ln|x+2|+\frac74\ln|x-2|=\Bigl(1-\frac34\Bigr)\ln|x+2|+=\Bigl(1+\frac34\Bigr)\ln|x-2|=\dots$$