समाकल int frac{sqrt{x^2+1} [log(x^2+1) - 2 lo | vedclass

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

Question

(A) माना $I = \int \frac{\sqrt{x^2+1} [\log(x^2+1) - 2 \log x]}{x^4} dx = \int \frac{x \sqrt{1+1/x^2} \log((1+1/x^2))}{x^4} dx = \int \frac{\sqrt{1+1/

Vedclass · Accepted Answer

$(1+\frac{1}{x^2})^{3/2} [-\frac{1}{3} \log(1+\frac{1}{x^2}) + \frac{2}{9}] + c$

Vedclass · Answer

(A) माना $I = \int \frac{\sqrt{x^2+1} [\log(x^2+1) - 2 \log x]}{x^4} dx = \int \frac{x \sqrt{1+1/x^2} \log((1+1/x^2))}{x^4} dx = \int \frac{\sqrt{1+1/x^2} \log(1+1/x^2)}{x^3} dx$.माना $t = 1+1/x^2$ है। तब $dt = -2/x^3 dx$, जिसका अर्थ है $dx/x^3 = -dt/2$.इन्हें समाकल में प्रतिस्थापित करने पर: $I = \int \sqrt{t} \log(t) (-1/2) dt = -1/2 \int t^{1/2} \log(t) dt$.खंडशः समाकलन का उपयोग करने पर: $\int u dv = uv - \int v du$, जहाँ $u = \log(t)$ और $dv = t^{1/2} dt$ है।$du = (1/t) dt$ और $v = (2/3) t^{3/2}$ है।$I = -1/2 [\log(t) \cdot (2/3) t^{3/2} - \int (2/3) t^{3/2} \cdot (1/t) dt] = -1/2 [\frac{2}{3} t^{3/2} \log(t) - \frac{2}{3} \int t^{1/2} dt]$.$I = -1/2 [\frac{2}{3} t^{3/2} \log(t) - \frac{2}{3} \cdot \frac{2}{3} t^{3/2}] + c = -\frac{1}{3} t^{3/2} \log(t) + \frac{2}{9} t^{3/2} + c$.$I = t^{3/2} [-\frac{1}{3} \log(t) + \frac{2}{9}] + c = (1+\frac{1}{x^2})^{3/2} [-\frac{1}{3} \log(1+\frac{1}{x^2}) + \frac{2}{9}] + c$.

समाकल $\int \frac{\sqrt{x^2+1} [\log(x^2+1) - 2 \log x]}{x^4} dx$ का मान किसके बराबर है...

Similar Questions

$\int \sqrt{x^{2}-8 x+7} \, dx$ का मान ज्ञात कीजिए।

$\int \frac{2 \cos 3 x-3 \sin 3 x}{\cos 3 x+2 \sin 3 x} d x=$

$0 < x < 1$ के लिए,समाकलन $\int [\operatorname{Tan}^{-1}(1-x+x^2) + \operatorname{Tan}^{-1}(1-x)] dx$ का मान ज्ञात कीजिए।

$\int \frac{1}{x^5 \sqrt[5]{x^5+1}} d x=$

$\int \frac{dx}{(1+x) \sqrt{8+7x-x^2}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module