Question

(tan² x + 2 tan x + 5) $\frac{dy}{dx}=2$ (1 + tan x) sec² x

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left({\tan }^{2}x+2\tan x+5\right)\frac{dy}{dx}=2\left(1+\tan x\right){\sec }^{2}x \\ \Rightarrow dy=\frac{2\left(1+\tan x\right){\sec }^{2}x}{\left({\tan }^{2}x+2\tan x+5\right)}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int \frac{2\left(1+\tan x\right){\sec }^{2}x}{\left({\tan }^{2}x+2\tan x+5\right)}dx.....\left(1\right) \\ Putting{\tan }^{2}x+2\tan x+5=t \\ \therefore \left(2\tan xse{c}^{2}x+2se{c}^{2}x\right)dx=dt \\ \Rightarrow 2\left(1+\tan x\right){\sec }^{2}xdx=dt \\ \mathrm{Therefore}\left(1\right)\mathrm{becomes}, \\ \int dy=\int \frac{1}{t}dt \\ \Rightarrow y=\log \left|t\right|+C \\ \Rightarrow y=\log \left|{\tan }^{2}x+2\tan x+5\right|+C \end{gathered}$$