<s>[INST] Integration By Parts (Logarithm) $$\int(2x+3)\ln (x)dx$$
My attempts, 
$$=\int(2x\ln (x)+3\ln (x))dx$$
$$=2\int x\ln (x)dx+3\int \ln(x)dx$$
For $x\ln (x)$, integrate by parts,then I got
$$=x^2\ln (x)-\int (x) dx+3\int \ln(x)dx$$
$$=x^2\ln (x)-\frac{x^2}{2}+3\int \ln(x)dx$$
For $\ln(x)$, integrate by parts, then I got 
$$=x^2\ln (x)-\frac{x^2}{2}+3x\ln (x)-3\int 1dx$$
$$=x^2\ln (x)-\frac{x^2}{2}+3x\ln (x)-3x+c$$
$$=\frac{1}{2}x(-x+2(x+3)\ln (x)-6)+c$$
But the given answer in book is $x^2\ln (x)-\frac{x^2}{2}+\frac{3}{x}+c$. What did I do wrong?
 [/INST] Bookish answer is wrong:
$$\left(x^2\ln (x)-\frac{x^2}{2}+\frac{3}{x}+c\right)'=2x\log x-\frac3{x^2}$$
And yours:
$$\left(\frac{1}{2}x(-x+2(x+3)\ln (x)-6)+c\right)'=2x\log x+3\log x=(2x+3)\log x$$
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