15. - / 1.07 points evaluate the integral. (use c for the constant of integration.) $int (ln(x))^{2} dx$

Explanation:

Step1: Use integration - by - parts formula $\int u;dv=uv-\int v;du$

Let $u = (\ln x)^2$, $dv=dx$. Then $du = 2\frac{\ln x}{x}dx$, $v = x$. So $\int(\ln x)^2dx=x(\ln x)^2-\int x\cdot2\frac{\ln x}{x}dx=x(\ln x)^2 - 2\int\ln x;dx$.

Step2: Use integration - by - parts on $\int\ln x;dx$

Let $u=\ln x$, $dv = dx$. Then $du=\frac{1}{x}dx$, $v = x$. So $\int\ln x;dx=x\ln x-\int x\cdot\frac{1}{x}dx=x\ln x - x$.

Step3: Substitute $\int\ln x;dx$ back

$\int(\ln x)^2dx=x(\ln x)^2-2(x\ln x - x)+C=x(\ln x)^2-2x\ln x + 2x+C$.

Answer:

$x(\ln x)^2-2x\ln x + 2x+C$