Question

Find the indefinite integral:
(t + 1) / (t^2 + 4)
Show steps.

Answer 1

Let t = 2tan x
dt/dx = 2 sec^2 x
dt = 2 sec^2 x dx
Therefore
\begin{eqnarray*}
\int{\frac{t+1}{t^2+4}}dt&=&\int{\frac{2\tan x + 1}{4\tan^2 x + 4}2\sec^2 x} dx \\
&=&\int{\frac{2\tan x + 1}{4\sec^2 x}2\sec^2 x}dx \\
&=&\int{\frac{2\tan x + 1}{2}} dx
\end{eqnarray*}
which should then be easy to integrate.

Answer 2

Sweet trig, thanks!

Answer 3

\[\int\frac{t+1}{t^2+4}dt=\int\frac{t}{t^2+4}dt+\int\frac{1}{t^2+4}dt=\]\[=\frac{1}{2}\int\frac{d(t^2+4)}{t^2+4}+\frac{1}{4}\int\frac{1}{(t/2)^2+1}dt=\frac{1}{2}\ln{(t^2+4)}+\frac{1}{2}\arctan\frac{t}{2}+C\]