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

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

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(A) Let $I = \int x^2 \cdot \frac{x \sec^2 x+	an x}{(x 	an x+1)^2} dx$. Using integration by parts, let $u = x^2$ and $dv = \frac{x \sec^2 x+	an x}{(x 	an x+1)^2} dx$. Then $du = 2x dx$ and $v = \int \frac{d(x 	an x)}{(x 	an x+1)^2} = -\frac{1}{x 	an x+1}$. So, $I = x^2 \left(-\frac{1}{x 	an x+1}ight) - \int 2x \left(-\frac{1}{x 	an x+1}ight) dx$. $I = -\frac{x^2}{x 	an x+1} + 2 \int \frac{x}{x 	an x+1} dx$. Multiply numerator and denominator by $\cos x$: $I = -\frac{x^2}{x 	an x+1} + 2 \int \frac{x \cos x}{x \sin x + \cos x} dx$. Let $t = x \sin x + \cos x$, then $dt = (x \cos x + \sin x - \sin x) dx = x \cos x dx$. $I = -\frac{x^2}{x 	an x+1} + 2 \int \frac{dt}{t} = -\frac{x^2}{x 	an x+1} + 2 \log|x \sin x + \cos x| + C$. Comparing with the given form, $A = 2$, $B = -1$, and $f(x) = x^2$. Thus, $f(A+B) = f(2-1) = f(1) = (1)^2 = 1$.

If $\int \frac{x^2(x \sec^2 x+\tan x)}{(x \tan x+1)^2} dx = A \log(|x \sin x+\cos x|) + B \frac{f(x)}{(x \tan x+1)} + C$, then $f(A+B) =$

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