समाकलन int_{-1}^2 log _eleft(x+sqrt{x^2+1}rig | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) माना $I = \int_{-1}^2 1 \cdot \log _e(x+\sqrt{x^2+1}) dx$.खंडशः समाकलन का उपयोग करते हुए,$\int u dv = uv - \int v du$,जहाँ $u = \log _e(x+\sqrt{x^

Vedclass · Accepted Answer

$\sqrt{2}-\sqrt{5}+\log _{e}\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$

Vedclass · Answer

(B) माना $I = \int_{-1}^2 1 \cdot \log _e(x+\sqrt{x^2+1}) dx$.खंडशः समाकलन का उपयोग करते हुए,$\int u dv = uv - \int v du$,जहाँ $u = \log _e(x+\sqrt{x^2+1})$ और $dv = dx$.तब $du = \frac{1}{x+\sqrt{x^2+1}} \cdot (1 + \frac{x}{\sqrt{x^2+1}}) dx = \frac{1}{\sqrt{x^2+1}} dx$ और $v = x$.$I = [x \log _e(x+\sqrt{x^2+1})]_{-1}^2 - \int_{-1}^2 \frac{x}{\sqrt{x^2+1}} dx$.$I = [x \log _e(x+\sqrt{x^2+1}) - \sqrt{x^2+1}]_{-1}^2$.$I = (2 \log _e(2+\sqrt{5}) - \sqrt{5}) - (-1 \log _e(-1+\sqrt{2}) - \sqrt{2})$.$I = 2 \log _e(2+\sqrt{5}) + \log _e(\sqrt{2}-1) - \sqrt{5} + \sqrt{2}$.चूँकि $\log _e(\sqrt{2}-1) = \log _e(\frac{1}{\sqrt{2}+1}) = -\log _e(\sqrt{2}+1)$,इसलिए:$I = \log _e(2+\sqrt{5})^2 - \log _e(\sqrt{2}+1) - \sqrt{5} + \sqrt{2}$.$I = \sqrt{2} - \sqrt{5} + \log _e\left(\frac{9+4\sqrt{5}}{\sqrt{2}+1}\right)$.

समाकलन $\int_{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x$ का मान है:

Similar Questions

यदि $f(x) = \begin{cases} 4x + 3, & 1 \le x \le 2 \\ 3x + 5, & 2 < x \le 4 \end{cases}$ है,तो $\int_1^4 f(x) \, dx = $

निश्चित समाकलन का मान ज्ञात कीजिए: $\int_{0}^{1} \left(1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \cdots \infty\right) e^{2x} \, dx$.

$\int_0^{\frac{\pi}{2}} x^3 \sin x \, dx =$

$\int_{0}^{20\pi} (\sin^4 x + \cos^4 x) dx$ का मान किसके बराबर है?

$\int_{-2}^{2} |1 - x^2| \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module