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Indefinite Integrals question

2023 · 6 Apr · Shift 1 · Q30
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  5. /2023 · 6 Apr · Shift 1 · Q30

Indefinite Integrals question

2023 · 6 Apr · Shift 1 · Q30

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
Let I(x)=∫x2(xsec⁡2x+tan⁡x)(xtan⁡x+1)2dxI(x)=\int \frac{x^{2}\left(x \sec ^{2} x+\tan x\right)}{(x \tan x+1)^{2}} d xI(x)=∫(xtanx+1)2x2(xsec2x+tanx)​dx. If I(0)=0I(0)=0I(0)=0, then I(π4)I\left(\frac{\pi}{4}\right)I(4π​) is equal to :
  1. A
    log⁡e(π+4)232−π24(π+4)\log _{e} \frac{(\pi+4)^{2}}{32}-\frac{\pi^{2}}{4(\pi+4)}loge​32(π+4)2​−4(π+4)π2​
  2. B
    log⁡e(π+4)216−π24(π+4)\log _{e} \frac{(\pi+4)^{2}}{16}-\frac{\pi^{2}}{4(\pi+4)}loge​16(π+4)2​−4(π+4)π2​
  3. C
    log⁡e(π+4)216+π24(π+4)\log _{e} \frac{(\pi+4)^{2}}{16}+\frac{\pi^{2}}{4(\pi+4)}loge​16(π+4)2​+4(π+4)π2​
  4. D
    log⁡e(π+4)232+π24(π+4)\log _{e} \frac{(\pi+4)^{2}}{32}+\frac{\pi^{2}}{4(\pi+4)}loge​32(π+4)2​+4(π+4)π2​
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