Evaluate the integral: int frac{2x+3}{sqrt{3x | vedclass

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(A) To solve $\int \frac{2x+3}{\sqrt{3x^2-2x+1}} dx$, we express the numerator as a multiple of the derivative of the quadratic expression inside the

Vedclass · Accepted Answer

$\frac{2}{3} \sqrt{3x^2-2x+1} + \frac{11}{3\sqrt{6}} \sinh^{-1}\left(\frac{3x-1}{\sqrt{2}}\right) + C$

Vedclass · Answer

(A) To solve $\int \frac{2x+3}{\sqrt{3x^2-2x+1}} dx$, we express the numerator as a multiple of the derivative of the quadratic expression inside the square root. The derivative of $3x^2-2x+1$ is $6x-2$. We write $2x+3 = \frac{1}{3}(6x-2) + \frac{11}{3}$. Thus, the integral becomes: $\int \frac{2x+3}{\sqrt{3x^2-2x+1}} dx = \frac{1}{3} \int \frac{6x-2}{\sqrt{3x^2-2x+1}} dx + \frac{11}{3} \int \frac{dx}{\sqrt{3(x^2 - \frac{2}{3}x + \frac{1}{3})}}$. For the first part, let $u = 3x^2-2x+1$, then $du = (6x-2)dx$. The integral is $\frac{1}{3} \int u^{-1/2} du = \frac{2}{3} \sqrt{3x^2-2x+1}$. For the second part, complete the square: $x^2 - \frac{2}{3}x + \frac{1}{3} = (x-\frac{1}{3})^2 + \frac{2}{9}$. So, $\frac{11}{3\sqrt{3}} \int \frac{dx}{\sqrt{(x-\frac{1}{3})^2 + (\frac{\sqrt{2}}{3})^2}} = \frac{11}{3\sqrt{3}} \sinh^{-1}\left(\frac{x-1/3}{\sqrt{2}/3}\right) = \frac{11}{3\sqrt{3}} \sinh^{-1}\left(\frac{3x-1}{\sqrt{2}}\right)$. Combining these, the result is $\frac{2}{3} \sqrt{3x^2-2x+1} + \frac{11}{3\sqrt{3}} \sinh^{-1}\left(\frac{3x-1}{\sqrt{2}}\right) + C$.

Evaluate the integral: $\int \frac{2x+3}{\sqrt{3x^2-2x+1}} dx$

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