Hello, spdrmncoo!
An interesting problem . . . it can be done
without Partial Fractions.
\(\displaystyle \L\int \frac{2x^2\,+\,x\,+\,7}{(x^2\,+\,4)^2}\,dx\)
Let
x=2tanθ⇒dx=2sec2θdθ
The numerator is:
2x2+x+7=8tan2θ+2tanθ+7
The denominator is: \(\displaystyle \,(x^2\,+\,4)^2\;=\;(4\tan^2\theta\,+\,4)^2\;=\

4\sec^2\theta)^2\;=\;16\sec^4\theta\)
The integral becomes: \(\displaystyle \L\,\int\frac{8\tan^2\theta\,+\,2\tan\theta\,+\,7}{16\sec^4\theta}\)\(\displaystyle \cdot(2\sec^2\theta\,d\theta) \;=\;\L\frac{1}{8}\int\frac{8\tan^2\theta\,+\,2\tan\theta\,+\,7}{\sec^2\theta}\ d\theta\)
Multiply top and bottom by \(\displaystyle \cos^2\theta:\L\;\frac{1}{8}\int\left(8\sin^2\theta\,+\,2\sin\theta\cos\theta\,+\,7\cos^2\theta\right)\ d\theta\)
We have: \(\displaystyle \L\:\frac{1}{8}\int\left(\sin^2\theta\,+\,2\sin\theta\cos\theta\,+\,7\sin^2\theta\,+\,7\cos^2\theta\right)\ d\theta\)
. . . . . \(\displaystyle \L=\;\frac{1}{8}\int\left(\sin^2\theta\,+\,2\sin\theta\cos\theta\,+\,7\right)\ d\theta\)
Hence: \(\displaystyle \L\:\frac{1}{8}\int \left(\frac{1\,-\,\cos2\theta}{2}\,+\,\sin2\theta\,+\,7\right)\ d\theta\;\) . . . etc.