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

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(A) Let $I = \int \frac{x^2(x \sec^2 x + 	an x)}{(x 	an x + 1)^2} dx$. Using integration by parts, let $u = x^2$ and $dv = \frac{x \sec^2 x + 	an x

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{x^2(x \sec^2 x + 	an x)}{(x 	an x + 1)^2} dx$. Using integration by parts, let $u = x^2$ and $dv = \frac{x \sec^2 x + 	an x}{(x 	an x + 1)^2} dx$. First, find $v = \int \frac{x \sec^2 x + 	an x}{(x 	an x + 1)^2} dx$. Let $t = x 	an x + 1$, then $dt = (x \sec^2 x + 	an x) dx$. So, $v = \int \frac{1}{t^2} dt = -\frac{1}{t} = -\frac{1}{x 	an x + 1}$. Now, applying the integration by parts formula $\int u dv = uv - \int v du$: $I = x^2 \left( -\frac{1}{x 	an x + 1} ight) - \int \left( -\frac{1}{x 	an x + 1} ight) (2x) dx$. $I = -\frac{x^2}{x 	an x + 1} + \int \frac{2x}{x 	an x + 1} dx$. $I = -\frac{x^2}{x 	an x + 1} + \int \frac{2x \cos x}{x \sin x + \cos x} dx$. Comparing this with the given expression, $f(x) = \int \frac{2x \cos x}{x \sin x + \cos x} dx$. Let $u = x \sin x + \cos x$, then $du = (x \cos x + \sin x - \sin x) dx = x \cos x dx$. Thus, $f(x) = \int \frac{2}{u} du = 2 \log |u| + c = 2 \log |x \sin x + \cos x| + c$.

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

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