$I=\int \frac{{{\sec }^{2}}x}{{{\tan }^{3}}+4\tan x}dx$

$I=\int \frac{{{\sec }^{2}}x}{{{\tan }^{3}}+4\tan x}dx$

Putting $t=tanx$

$dt={{\sec }^{2}}xdx$

$I=\int \frac{dt}{({{t}^{3}}+4)}=\int \frac{dt}{t\left( {{t}^{2}}+4 \right)}$

Now, putting

$\frac{1}{t\left( {{t}^{2}}+4 \right)}=\frac{A}{t}+\frac{Bt+C}{{{t}^{2}}+4}.....(1)$

$A\left( {{t}^{2}}+4 \right)+(Bt+C)t=1$

Putting $t=0$

$A\left( 0+4 \right)\times B\left( 0 \right)=1$
$A=\frac{1}{4}$

By equating the coefficients of ${{t}^{2}}$ and constant here,

$A+B=0$

$\frac{1}{4}+B=0$

$B=-\frac{1}{4},C=0$

Now from equation (1) we get,

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

$=\frac{1}{4}\log t-\frac{1}{4}\times \frac{1}{2}\log ({{t}^{2}}+4)+c$

$=\frac{1}{4}\log \tan x-\frac{1}{8}\log \left( {{\tan }^{2}}x+4 \right)+c$

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