A definite integral involve Logarithmic Functions Here is the integral body:
$$\int_0^m {{x^a}\ln \left( {x + b} \right)dx,a >  - \frac{3}{2},b > 0} $$
 A: As mentioned in the comments, integration by parts is a good place to start. The $a= -1$ case has to be handled separately (and it's not difficult to show the integral diverges in this case), but otherwise,
$$I(m,a,b)=\int_{0}^{m}x^a\log{\left(x+b\right)}dx=\frac{1}{a+1}x^{a+1}\log{\left(x+b\right)}\big|_{0}^{m}-\frac{1}{a+1}\int_{0}^{m}\frac{x^{a+1}}{x+b}dx\\
=\frac{m^{a+1}\log{\left(m+b\right)}}{a+1}-\frac{1}{a+1}\int_{0}^{m}\frac{x^{a+1}}{x+b}dx.$$
Since by assumption $b>0$, it's a good candidate for a rescaling argument. Substituting $x=bu$ into the integral that resulted from IBPs abov: $~~dx=b\,du$, $u=\frac{x}{b}$, and so
$$\int_{0}^{m}\frac{x^{a+1}}{x+b}dx=\int_{0}^{\frac{m}{b}}\frac{(bu)^{a+1}}{bu+b}(b\,du)\\
=b^{a+1}\int_{0}^{\frac{m}{b}}\frac{u^{a+1}}{u+1}\,du.$$
Simplifications past this this point will depend a great deal upon the individual's opinion about what constitutes 'simpler'.
A: $\displaystyle\int_0^m\frac{x^{n-1}}{x+a}dx=\frac{m^n}{a\cdot n}\cdot~_{_2}F_{_1}\bigg(1,n~;~n+1~;-\frac ma\bigg),$ where $_{_2}F_{_1}$ is the Gaussian hypergeometric function. For large values of m, we have $\displaystyle\int_0^\infty\frac{x^n}{x+a}dx=-a^n\cdot\pi\cdot\csc(n\pi)$.
