11) ∫₀¹ 2x / (x² - 3) dx

Explanation:

Step1: Use substitution

Let $u = x^{2}-3$, then $du = 2x dx$. When $x = 0$, $u=-3$; when $x = 1$, $u=-2$.

Step2: Rewrite the integral

The integral $\int_{0}^{1}\frac{2x}{x^{2}-3}dx$ becomes $\int_{-3}^{-2}\frac{du}{u}$.

Step3: Integrate

The antiderivative of $\frac{1}{u}$ is $\ln|u|$. So $\int_{-3}^{-2}\frac{du}{u}=\left[\ln|u|\right]_{-3}^{-2}$.

Step4: Evaluate the definite - integral

$\left[\ln|u|\right]_{-3}^{-2}=\ln|-2|-\ln|-3|=\ln 2-\ln 3=\ln\frac{2}{3}$.

Answer:

$\ln\frac{2}{3}$