A question about my solution of this integral $\int_{0}^{1} \frac{a^x-a}{a^x+a} dx$ So i had to solve this integral:
$$\int_{0}^{1} \frac{a^x-a}{a^x+a} dx$$
$a\in\mathbb{R^+}$
So first i used substitution:
$t=a^x+a \implies a^x=t-a $ 
$ dx= \frac{dt}{\ln(a)(t-a)}$
Then with partial fractions i got this:
$$\frac{1}{\ln a} \left(2 \int_{1+a}^{2a}\frac{1}{t}dt-\int_{1+a}^{2a}\frac{1}{t-a}dt \right)$$
And with that i just used the common integral and have  gotten this solution afterwards:
$$\frac{\ln(2a(1+a)^2)}{\ln a}$$
So i wonder if i did it right, at most the problem is with substitution but i think i did change bounds correctly, but any insight would be helpful.
Thank you in advance.

hint: setting $$a^x=t+a$$ then we get$$x\ln(a)=t+a$$ and $$dx=\frac{1}{\ln(a)}dt$$
