4. -/6.25 points details my notes evaluate the integral. (use c for the constant of integration. ∫(ln(x))²…

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, $\int(\ln x)^2dx=x(\ln x)^2-\int x\cdot2\frac{\ln x}{x}dx=x(\ln x)^2 - 2\int\ln xdx$.

Step2: Use integration - by - parts again for $\int\ln xdx$

Let $u=\ln x$, $dv = dx$. Then $du=\frac{1}{x}dx$, $v = x$. So, $\int\ln xdx=x\ln x-\int x\cdot\frac{1}{x}dx=x\ln x - x+C_1$.

Step3: Substitute $\int\ln xdx$ back into the previous result

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

Answer:

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