\(\rm \int x\tan^{-1}x dx\) का मूल्यांकन कीजिए। 

  1. \(\rm {1\over2}\left[x^2\tan^{-1}x - x\right]+c\)
  2. \(\rm {1\over2}\left[(x^2+x)\tan^{-1}x - x\right]+c\)
  3. \(\rm {1\over2}\left[(x^2+1)\tan^{-1}x - (x + 1)\right]+c\)
  4. \(\rm {1\over2}\left[(x^2+1)\tan^{-1}x - x\right]+c\)

Answer (Detailed Solution Below)

Option 4 : \(\rm {1\over2}\left[(x^2+1)\tan^{-1}x - x\right]+c\)
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NDA 01/2025: English Subject Test
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30 Questions 120 Marks 30 Mins

Detailed Solution

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संकल्पना:

खंडशः समाकलन: खंडशः समाकलन गुणनफलों का समाकलन ज्ञात करने की एक विधि है। 

खंडशः समाकलन के लिए सूत्र को निम्न द्वारा ज्ञात किया गया है:

⇒ \(\rm ∫ u vdx=u ∫ vdx- ∫ \left({du\over dx}\times ∫ vdx\right)dx \) + C

जहाँ u फलन u(x) है और v फलन v(x) है। 

ILATE नियम में विशेष रूप से इस नियम की वरीयता क्रम व्युत्क्रम, लघुगुणक, बीजगणितीय, त्रिकोणमितीय और घातांक जैसे कुछ फलनों पर आधारित होती है। 

गणना:

I = \(\rm \int x\tan^{-1}x dx\)

I = \(\rm \tan^{-1}x\int xdx - \int \left({1\over1+x^2}\int xdx\right)dx+c\)

I = \(\rm {x^2\tan^{-1}x\over2} - {1\over2}\int \left({x^2\over1+x^2}\right)dx+c\)

I = \(\rm {x^2\tan^{-1}x\over2} - {1\over2}\int \left(1-{1\over1+x^2}\right)dx+c\)

I = \(\rm {1\over2}\left[x^2\tan^{-1}x - (x-\tan^{-1}x)\right]+c\)

I = \(\boldsymbol{\rm {1\over2}\left[(x^2+1)\tan^{-1}x - x\right]+c}\)

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