How should I simplify this $\tan^{-1}$ expression? Integral question I have to integrate this function 
$$I = \int_0^4\frac{20x-5x^2}{x^2+9} \mathrm{d}x$$
obtaining 
$$20\ln(5/3) + 15\tan^{-1}(4/3) -20.$$
However, my calculator, even after somehow simplifying it a bit, gives this: 
$$\frac{40\ln(5/3) -30\tan^{-1}(3/4) +15\pi -40}{2}$$
As you can see there is something wrong with arctan integration, can anybody help or know how to simplify this with some identity? 
I am asking this because it is one of the questions in the practise exam, and in the exam i will have to use this calculator, no others allowed, so if a question like this comes up, i will get stuck..
 A: The answers are the same when you notice that, for $x\gt0$,
$$
\tan^{-1}\left(\frac1x\right)=\frac\pi2-\tan^{-1}(x)
$$

Note that
$$
\tan^{-1}\left(\frac ba\right)=B\quad\text{and}\quad\tan^{-1}\left(\frac ab\right)=A
$$
and
$$
A+B=\frac\pi2
$$
$\hspace{3.4cm}$

If you want to use trigonometric identities,
$$
\begin{align}
\tan\left(\frac\pi2-x\right)
&=\frac{\sin\left(\frac\pi2-x\right)}{\cos\left(\frac\pi2-x\right)}\\
&=\frac{\sin\left(\frac\pi2\right)\cos(x)-\cos\left(\frac\pi2\right)\sin(x)}{\cos\left(\frac\pi2\right)\cos(x)+\sin\left(\frac\pi2\right)\sin(x)}\\[6pt]
&=\frac{\cos(x)}{\sin(x)}\\[12pt]
&=\frac1{\tan(x)}
\end{align}
$$
A: By first simplifying the fraction, we get that $$\frac{40\ln(5/3) -30\tan^{-1}(3/4) +15\pi -40}{2} = 20\ln(5/3) -15\tan^{-1}(3/4) +\frac{15}{2}\pi -20$$
Subtracting $20\ln(5/3) + 15\tan^{-1}(4/3) -20$ from it, we get $$(20\ln(5/3) -15\tan^{-1}(3/4) +\frac{15}{2}\pi -20) - (20\ln(5/3) + 15\tan^{-1}(4/3) -20)$$
$$ = \frac{15}{2}\pi - 15\tan^{-1}(4/3) -15\tan^{-1}(3/4) = 15 \left(\frac{\pi}{2} - \tan^{-1} (4/3) - \tan^{-1} (3/4)\right)$$
Here we bring in the definition of tangent. Given a right triangle $ABC$ with $C$ being the right angle, $$\tan \angle ABC = \frac{AC}{BC}$$ From this, we get that $$\angle ABC = \tan^{-1} \frac{AC}{BC}$$
Now, we take a $3-4-5$ triangle and note that the two inverse tangents are the two angles that are not $\displaystyle \frac{\pi}{2}$. Because the angles of a triangle add up to $\pi$, we note that they sum to $\displaystyle \frac{\pi}{2}$. Therefore, $$15 \left(\frac{\pi}{2} - \tan^{-1} (4/3) - \tan^{-1} (3/4)\right) = 15 \left(\frac{\pi}{2} - \frac{\pi}{2}\right) = 0$$
Q.E.D.
