int frac{(3 x-2) tan left(sqrt{9 x^2-12 x+1}r | vedclass

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(D) Let $I = \int \frac{(3 x-2) 	an \left(\sqrt{9 x^2-12 x+1}ight)}{\sqrt{9 x^2-12 x+1}} d x$. Let $u = \sqrt{9 x^2-12 x+1}$. Then $u^2 = 9 x^2-12

Vedclass · Accepted Answer

$\frac{1}{3} \log \left|\sec \sqrt{9 x^2-12 x+1}\right|+c$

Vedclass · Answer

(D) Let $I = \int \frac{(3 x-2) 	an \left(\sqrt{9 x^2-12 x+1}ight)}{\sqrt{9 x^2-12 x+1}} d x$. Let $u = \sqrt{9 x^2-12 x+1}$. Then $u^2 = 9 x^2-12 x+1$. Differentiating both sides with respect to $x$,we get $2u \frac{du}{dx} = 18x - 12 = 6(3x - 2)$. Thus,$u \frac{du}{dx} = 3(3x - 2)$,which implies $(3x - 2) dx = \frac{1}{3} u du$. Substituting these into the integral,we get $I = \int \frac{	an(u)}{u} \cdot \frac{1}{3} u du = \frac{1}{3} \int 	an(u) du$. The integral of $	an(u)$ is $\log |\sec(u)| + c$. Therefore,$I = \frac{1}{3} \log |\sec(\sqrt{9 x^2-12 x+1})| + c$.

$\int \frac{(3 x-2) \tan \left(\sqrt{9 x^2-12 x+1}\right)}{\sqrt{9 x^2-12 x+1}} d x=$

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