Solve the indefinite integral $\int\frac{\ln^2(a+bx)}{x^n}$

Question

The given integral $\int\frac{\ln^2(a+bx)}{x^n}$ solve for the parcial method. I didn't know. Please anyone help me. I now how to solve the $\int\frac{\ln(a+bx)}{x^n},$ but didn't know how to find the $\int\frac{\ln^2(a+bx)}{x^n}.$

I tried: $\int\frac{\ln^2(a+bx)}{x^n}dx=\frac{\ln^2(a+bx)}{x^{n−1}(1−n)}−\frac{2b}{1−n}∫\frac{\ln(a+bx)}{x^{n−1}(a+bx)}dx,$ but now i didn't know how to continue. How to solve the integral $$∫\frac{\ln(a+bx)}{x^{n−1}(a+bx)}dx$$

Answer 1

Use integration by parts $$\int\frac{\ln^2(a+bx)}{x^n}dx= \frac{\ln^2(a+bx)}{x^{n-1}(1-n)}-\frac{2b}{1-n}\int \frac{\ln(a+bx)}{x^{n-1}(a+bx)} \,dx $$

You can then use partial fraction decomposition.

Note that

$$\frac{1}{x^n(a+bx)} = \frac{a+bx-bx}{ax^n(a+bx)} = \frac{1}{ax^n}-\frac{b}{ax^{n-1}(a+bx)}$$

Continue like this until

$$\frac{1}{x(a+bx)} = \frac{1}{ax}-\frac{b}{a(a+bx)}$$