Question

(17) $\int x\cos xdx =$
(18) $\int x\ln xdx =$

Explanation:

Step1: Use integration by parts for $\int x\cos xdx$

Integration by parts formula: $\int u dv=uv-\int v du$. Let $u = x$, $dv=\cos xdx$. Then $du = dx$, $v=\sin x$.
$$\int x\cos xdx=x\sin x-\int\sin xdx$$

Step2: Integrate $\int\sin xdx$

We know that $\int\sin xdx=-\cos x + C$.
$$x\sin x-\int\sin xdx=x\sin x+\cos x + C$$

Step3: Use integration by parts for $\int x\ln xdx$

Let $u=\ln x$, $dv = xdx$. Then $du=\frac{1}{x}dx$, $v=\frac{1}{2}x^{2}$.
$$\int x\ln xdx=\frac{1}{2}x^{2}\ln x-\int\frac{1}{2}x^{2}\cdot\frac{1}{x}dx$$

Step4: Simplify and integrate $\int\frac{1}{2}x dx$

Simplify $\int\frac{1}{2}x dx=\frac{1}{4}x^{2}+C$.
$$\frac{1}{2}x^{2}\ln x-\int\frac{1}{2}x dx=\frac{1}{2}x^{2}\ln x-\frac{1}{4}x^{2}+C$$

Answer:

For (17): $x\sin x+\cos x + C$; For (18): $\frac{1}{2}x^{2}\ln x-\frac{1}{4}x^{2}+C$