TS EAMCET · Maths · Indefinite Integration
\(\int\left(\frac{4 \tan ^4 x+3 \tan ^2 x-1}{\tan ^2 x+4}\right) d x=\)
- A \(4 \tan x-\frac{17}{4} \tan ^{-1}\left(\frac{\tan x}{4}\right)+c\)
- B \(4 \tan x-\frac{17}{4} \tan ^{-1}\left(\frac{\tan x}{2}\right)+c\)
- C \(4 \tan x-\frac{17}{2} \tan ^{-1}\left(\frac{\tan x}{2}\right)+c\)
- D \(2 \tan x-\frac{17}{2} \tan ^{-1}\left(\frac{\tan x}{2}\right)+c\)
Answer & Solution
Correct Answer
(C) \(4 \tan x-\frac{17}{2} \tan ^{-1}\left(\frac{\tan x}{2}\right)+c\)
Step-by-step Solution
Detailed explanation
\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…
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