Evaluate the value of x ⁻¹ x\, dx: / x ⁻¹ x\, dx का मान ज्ञात कीजिए:

(x²/2)⁻¹ x - (1/2)x + (1/2)⁻¹ x + C

-(x²/2)⁻¹ x + (1/2)x + (1/2)⁻¹ x + C

(x²/2)⁻¹ x + (1/2)x - (1/2)⁻¹ x + C

-(x²/2)⁻¹ x - (1/2)x - (1/2)⁻¹ x + C

(x²/2)⁻¹ x - (1/2)x + (1/2)⁻¹ x + C

Evaluate the value of x ⁻¹ x\, dx: / x ⁻¹ x\, dx का मान ज्ञात कीजिए:

(4) (x²/2)⁻¹ x - (1/2)x + (1/2)⁻¹ x + C Explanation (English): Integrate by parts using u\, v'\, dx = u v - u'\, v\, dx: Let u = ⁻¹ x u' = (1/1 + x²) Let v' = x v = (x²/2) x ⁻¹ x\, dx = (x²/2)⁻¹ x - (x²/2(1 + x²))\, dx = (x²/2)⁻¹ x - (1/2) ((x² + 1) - 1/1 + x²)\, dx = (x²/2)⁻¹ x - (1/2) (1 - (1/1 + x²))\, dx = (x²/2)⁻¹ x - (1/2)x + (1/2)⁻¹ x + C Hence, Option 4 is correct. व्याख्या (हिंदी): खंडशः समाकलन (Integration by parts) विधि द्वारा: u = ⁻¹ x तथा dv = x\, dx v = (x²/2): x ⁻¹ x\, dx = (x²/2)⁻¹ x - (1/2) (x²/1 + x²)\, dx = (x²/2)⁻¹ x - (1/2) (1 - (1/1 + x²))\, dx = (x²/2)⁻¹ x - (1/2)x + (1/2)⁻¹ x + C अतः विकल्प 4 सही उत्तर है।

Evaluate the value of x ⁻¹ x\, dx: - Solved MCQ

Q.103

ICAR AIEEA PG 2024 - Statistical Sciences

MCQ (Single Correct) +4 −1

Evaluate the value of $\int x \tan^{-1} x\, dx$:

$\int x \tan^{-1} x\, dx$ का मान ज्ञात कीजिए:

$-\frac{x^2}{2}\tan^{-1} x + \frac{1}{2}x + \frac{1}{2}\tan^{-1} x + C$ $-\frac{x^2}{2}\tan^{-1} x + \frac{1}{2}x + \frac{1}{2}\tan^{-1} x + C$ $\frac{x^2}{2}\tan^{-1} x + \frac{1}{2}x - \frac{1}{2}\tan^{-1} x + C$ $\frac{x^2}{2}\tan^{-1} x + \frac{1}{2}x - \frac{1}{2}\tan^{-1} x + C$ $-\frac{x^2}{2}\tan^{-1} x - \frac{1}{2}x - \frac{1}{2}\tan^{-1} x + C$ $-\frac{x^2}{2}\tan^{-1} x - \frac{1}{2}x - \frac{1}{2}\tan^{-1} x + C$ $\frac{x^2}{2}\tan^{-1} x - \frac{1}{2}x + \frac{1}{2}\tan^{-1} x + C$ $\frac{x^2}{2}\tan^{-1} x - \frac{1}{2}x + \frac{1}{2}\tan^{-1} x + C$ Correct

Integrate by parts using $\int u\, v'\, dx = u v - \int u'\, v\, dx$:
Let $u = \tan^{-1} x \implies u' = \frac{1}{1 + x^2}$
Let $v' = x \implies v = \frac{x^2}{2}$

$$\int x \tan^{-1} x\, dx = \frac{x^2}{2}\tan^{-1} x - \int \frac{x^2}{2(1 + x^2)}\, dx$$ $$= \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}\int \frac{(x^2 + 1) - 1}{1 + x^2}\, dx$$ $$= \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}\int \left(1 - \frac{1}{1 + x^2}\right)\, dx$$ $$= \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}x + \frac{1}{2}\tan^{-1} x + C$$

Hence, Option 4 is correct.

खंडशः समाकलन (Integration by parts) विधि द्वारा:
$u = \tan^{-1} x$ तथा $dv = x\, dx \implies v = \frac{x^2}{2}$:

$$\int x \tan^{-1} x\, dx = \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}\int \frac{x^2}{1 + x^2}\, dx$$ $$= \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}\int \left(1 - \frac{1}{1 + x^2}\right)\, dx$$ $$= \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}x + \frac{1}{2}\tan^{-1} x + C$$

अतः विकल्प 4 सही उत्तर है।

Explanation not matching this question? क्या व्याख्या इस प्रश्न से मेल नहीं खा रही? Report

Want to practice this paper in exam conditions? क्या आप परीक्षा जैसे माहौल में अभ्यास करना चाहते हैं? Test yourself in 5-question batch format with timer and hidden answers. 5-प्रश्नों के बैच फॉर्मैट, टाइमर और छिपे उत्तरों के साथ खुद को परखें।

Start Practice Mode → प्रैक्टिस टेस्ट मोड शुरू करें →