Evaluate the following integral:

Solution:

Re-writing the given equation as

Multiplying $\sec ^{4} x$ in both numerator and denominator

$\int \frac{\sec ^{4} x}{\tan ^{4} x+\tan ^{2} x+1} d x$

$=\int \frac{\left(\tan ^{2} x+1\right) \sec ^{2} x}{\tan ^{4} x+\tan ^{2} x+1} d x$

Assume $\tan x=t$

$\sec ^{2} x d x=d t$

$=\int \frac{\left(\mathrm{t}^{2}+1\right) \mathrm{dt}}{\mathrm{t}^{4}+\mathrm{t}^{2}+1}$

$=\int \frac{1+\frac{1}{\mathrm{t}^{2}}}{\mathrm{t}^{2}+1+\frac{1}{\mathrm{t}^{2}}} \mathrm{dt}$

$=\int \frac{1+\frac{1}{\mathrm{t}^{2}}}{\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)^{2}+3} \mathrm{dt}$

Assume $\mathrm{z}=\mathrm{t}-\frac{1}{\mathrm{t}}$

$\Rightarrow \mathrm{dz}=1+\frac{1}{\mathrm{t}^{2}}$

$=\int \frac{\mathrm{dz}}{\mathrm{z}^{2}+3}$

Using identity $\int \frac{1}{x^{2}+1} d x=\arctan (x)$

$=\frac{1}{\sqrt{3}} \arctan \left(\frac{\mathrm{z}}{\sqrt{3}}\right)+\mathrm{c}$

$=\frac{1}{\sqrt{3}} \arctan \left(\frac{\mathrm{t}-\frac{1}{\mathrm{t}}}{\sqrt{3}}\right)+\mathrm{c}$

$=\frac{1}{\sqrt{3}} \arctan \left(\frac{\tan x-\frac{1}{\tan x}}{\sqrt{3}}\right)+c$