Q113 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
$\int\left(\frac{4 \tan ^4 x+3 \tan ^2 x-1}{\tan ^2 x+4}\right) d x=$
- A$4 \tan x-\frac{17}{4} \tan ^{-1}\left(\frac{\tan x}{4}\right)+c$
- B$4 \tan x-\frac{17}{4} \tan ^{-1}\left(\frac{\tan x}{2}\right)+c$
- Correct answerC$4 \tan x-\frac{17}{2} \tan ^{-1}\left(\frac{\tan x}{2}\right)+c$
- D$2 \tan x-\frac{17}{2} \tan ^{-1}\left(\frac{\tan x}{2}\right)+c$
Solution
$\begin{aligned} & \text { } \quad I=\int \frac{4 \tan ^4 x+3 \tan ^2 x-1}{\tan ^2 x+4} d x \\ & I=\int \frac{\left(4 \tan ^2 x-1\right)\left(\tan ^2 x+1\right)}{\tan ^2 x+4} d x \\ & I=\int \frac{\sec ^2 x(2 \tan x-1)(2 \tan x+1)}{\tan ^2 x+4} d x \\ & \text { Let } \tan x=u \Rightarrow \sec ^2 x d x=d u \\ & \Rightarrow I=\int \frac{(2 u-1)(2 u+1)}{u^2+4} d u=\int \frac{4\left(u^2+4\right)-17}{\left(u^2+4\right)} d u \\ & \Rightarrow I=\int\left(4-\frac{17}{u^2+4}\right) d u=4 \int d u-17 \int \frac{d u}{u^2+4} \\ & \Rightarrow I=4 u-\frac{17}{2} \tan ^{-1}\left(\frac{u}{2}\right)+C \\ & \Rightarrow I=4 \tan x-\frac{17}{2} \tan ^{-1}\left(\frac{\tan x}{2}\right)+C\end{aligned}$
Hence, the correct answer is option: 'C'.