Having trouble computing $\int_3^5\frac{t}{1+0.1t} dt $ $$\int_3^5\frac{t}{1+0.1t} dt $$
For some reason this is equal to: 

1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))

I have no idea how to reduce to that.
 A: Hint:
$$\frac{t}{1+0.1t} = \frac{10\cdot(1+0.1t) - 10}{1+0.1t} = 10 - \frac{10}{1+0.1t}$$
A: $$
\frac{x}{1+0.1x}=\frac{x}{1+0.1x}\cdot\frac{10}{10}=
\frac{10x}{10+x}=10\left(\frac{x}{10+x}\right)=\\
10\left(\frac{-10+10+x}{10+x}\right)=
10\left(\frac{-10}{10+x}+\frac{10+x}{10+x}\right)=
10\left(-\frac{10}{10+x}+1\right)=\\
10\left(1-\frac{10}{10+x}\right)=10-\frac{100}{10+x}.
$$

$$
\int\left(10-\frac{100}{10+x}\right)\,dx=
10\int\,dx-100\int\frac{1}{10+x}\frac{d}{dx}(10+x)\,dx=\\
10x-100\int\frac{1}{10+x}\,d(10+x)=
10x-100\ln{|10+x|}+C.
$$

$$
\int_3^5\frac{t}{1+0.1t}\,dt=
\bigg[10t-100\ln{|10+t|}\bigg]_3^5=\\
50-100\ln{15}-(30-100\ln{13})=
20-100\ln{15}+100\ln{13}=\\
20-100(\ln{15}-\ln{13})=20-100\ln{\frac{15}{13}}.
$$

The answer you gave is equivalent to what I got:
$$
\frac{1}{0.1}\left(2-\frac{1}{0.1}\left[\ln{1.5}-\ln{1.3}\right]\right)=
10\left(2-10\left[\ln{\frac{15}{10}}-\ln{\frac{13}{10}}\right]\right)=\\
20-100\ln{\left(\frac{15}{10}\div\frac{13}{10}\right)}=
20-100\ln{\frac{15}{13}}.
$$
