Let I(x)=\int \frac{x^2 (x \sec ^2 x+\tan x) }{(x \tan x+1)^2} \dx. If I(0)=0, then I (\frac{\pi}{4}) is equal to : [6-Apr-2023 shift 1]

Correct Answer is : loge(π+4)232−π24(π+4)

\log _e \frac{(\pi+4)^2}{16}+\frac{\pi^2}{4(\pi+4)}

\log _{ {\e}} \frac{(\pi+4)^2}{32}-\frac{\pi^2}{4(\pi+4)}

\log _e \frac{(\pi+4)^2}{16}-\frac{\pi^2}{4(\pi+4)}

\log _e \frac{(\pi+4)^2}{32}+\frac{\pi^2}{4(\pi+4)}

Indefinite Integrals

Section: Mathematics

Question : 29 of 89

Marks: +1, -0

Let I(x)=∫‌

x2(x sec2x+tan‌x)

(x‌tan‌x+1)2

‌dx. If I(0)=0, then I(‌

) is equal to :

[6-Apr-2023 shift 1]

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