यदि int frac{x^2(x sec^2 x+tan x)}{(x tan x+1 | vedclass

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(A) माना $I = \int \frac{x^2(x \sec^2 x + 	an x)}{(x 	an x + 1)^2} dx$. खंडशः समाकलन (integration by parts) का उपयोग करते हुए, $u = x^2$ और $dv = \f

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$2 \log |x \sin x + \cos x| + c$

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(A) माना $I = \int \frac{x^2(x \sec^2 x + 	an x)}{(x 	an x + 1)^2} dx$. खंडशः समाकलन (integration by parts) का उपयोग करते हुए, $u = x^2$ और $dv = \frac{x \sec^2 x + 	an x}{(x 	an x + 1)^2} dx$ लें। सबसे पहले, $v = \int \frac{x \sec^2 x + 	an x}{(x 	an x + 1)^2} dx$ ज्ञात करें। माना $t = x 	an x + 1$, तो $dt = (x \sec^2 x + 	an x) dx$। अतः, $v = \int \frac{1}{t^2} dt = -\frac{1}{t} = -\frac{1}{x 	an x + 1}$। अब, खंडशः समाकलन का सूत्र $\int u dv = uv - \int v du$ लागू करने पर: $I = x^2 \left( -\frac{1}{x 	an x + 1} ight) - \int \left( -\frac{1}{x 	an x + 1} ight) (2x) dx$। $I = -\frac{x^2}{x 	an x + 1} + \int \frac{2x}{x 	an x + 1} dx$। $I = -\frac{x^2}{x 	an x + 1} + \int \frac{2x \cos x}{x \sin x + \cos x} dx$। दिए गए व्यंजक के साथ तुलना करने पर, $f(x) = \int \frac{2x \cos x}{x \sin x + \cos x} dx$। माना $u = x \sin x + \cos x$, तो $du = (x \cos x + \sin x - \sin x) dx = x \cos x dx$। अतः, $f(x) = \int \frac{2}{u} du = 2 \log |u| + c = 2 \log |x \sin x + \cos x| + c$।

यदि $\int \frac{x^2(x \sec^2 x+\tan x)}{(x \tan x+1)^2} dx = \frac{-x^2}{x \tan x+1} + f(x) + c$ है, तो $f(x) =$

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