Complete the square in the denominator to get
tan²(t ) + 14 tan(t ) + 48 = tan²(t ) + 14 tan(t ) + 49 - 1
… = (tan(t ) + 7)² - 1
Substitute u = tan(t ) + 7 and du = sec²(t ) dt. Then the integral becomes
[tex]\displaystyle \int \frac{2\sec^2(t)}{\tan^2(t)+14\tan(t)+48} \,\mathrm dt = 2 \int \frac{\mathrm du}{u^2-1}[/tex]
Separate the integrand into partial fractions:
[tex]\dfrac1{u^2-1} = \dfrac12 \left(\dfrac1{u-1}-\dfrac1{u+1}\right)[/tex]
Then we get
[tex]\displaystyle \int \frac{2\sec^2(t)}{\tan^2(t)+14\tan(t)+48} \,\mathrm dt = \int \left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm du \\\\ =\ln|u-1|-\ln|u+1| + C \\\\ = \ln\left|\frac{u-1}{u+1}\right|+C \\\\ = \boxed{\ln\left|\frac{\tan(t)+6}{\tan(t)+8}\right|+C}[/tex]
