On page 188.

Lemma 10.2.3 is $\sum_{\substack{\delta \leq x \\ 2 \nmid \delta}}\frac{\mu^{2}(\delta)}{\phi_{1}(\delta)} = A_{1}log(x) +A_{2} + O(\frac{1}{x^{1/4}})$ for positive $A_{1}$, $A_{2}$.

And $I = \int_{3}^{x} (\sum_{\substack{\delta \leq x \\ 2 \nmid \delta}}\frac{\mu^{2}(\delta)}{\phi_{1}(\delta)}) \frac{1}{t (1+log(\frac{x}{t}))} dt$ 

Now my question is how did they get the following equation by lemma 10.2.3?

$I = -A_{1} \int_{3}^{x} \frac{dt}{t}  + (A_{1}log(x)+A_{1}+A_{2})\int_{3}^{x} \frac{dt}{t(1+log(x)-log(t)} + O(\frac{1}{x^{1/4}})$

I have been stuck on this for a long while. Any help is appreciated. :)

Edited