integral $\int(4x+9)\ln x\,dx$

Question

I have trouble calculating this integral. I include my calculations for easier troubleshooting.

$$\int(4x+9)\ln x\,dx = 4\int x \ln x \,dx + 9\int x \ln x\,dx = 4(x \ln x - x)+9(x \ln x - x)$$

Answer 1

You have a typo in the second term, as mentioned in a comment. Regarding the first term, integration by parts gives $$ \int x \ln x dx = \frac12 x^2 \ln x-\int \frac 12 x^2 \cdot \frac 1x dx = \frac 12 x^2 \ln x - \frac 14 x^2 + C. $$

Answer 2

using integration by parts

In the general case $$\int\;(ax＋b)\ln(x)dx＝(\frac {ax^2}{2}＋bx)\ln(x)－\int\;(\frac {ax^2}{2}＋bx)×\frac {1}{x}\;dx\\~\\＝(\frac {ax^2}{2}＋bx)\ln(x)－\int(\frac {ax}{2}＋b)\;dx＝(\frac {ax^2}{2}＋bx)\ln(x)－(\frac {ax^2}{4}＋bx)\\~\\＝\frac {ax^2}{2}(\ln(x)－\textstyle\frac {1}{2}))＋bx(\ln(x)－\textstyle\frac {1}{2})-\frac {bx}{2}\\~\\＝(\ln(x)－\textstyle\frac {1}{2}))(\frac {ax^2}{2}＋bx)－\frac {bx}{2}\\~\\$$

In your example we have a =4 and b=9 So : $$\int\;(4x＋9)\ln(x)dx＝(\ln(x)－\textstyle\frac {1}{2})(\frac {4x^2}{2}＋9x)－\frac {9x}{2}＝(\ln(x)－\textstyle\frac {1}{2})(2x^2＋9x)－\frac {9x}{2}$$