# How to solve the indefinite integral?

The integral :-

$$\int x^m \ln(a+x) \,dx.$$

(Also what is $$m$$ is not an integer, just an arbitrary real number?) I have found the integral in the book gradshteyn and ryzhik of which this is a special case. I tried integral by parts for $$a=0$$ it follows trivially but for the other case please help to find the integral?

• Perhaps you could do $u=a+x$ and then expand $x^m=(u-a)^m$ by the binomial theorem. – J_P Jun 21 at 18:28
• Oh yes thank you very much – Bijayan Ray Jun 21 at 18:30

Use integration by parts and note that $$\frac{x^{m+1}}{a+x}=\sum_{k=0}^m(-a)^kx^{m-k}+\frac{(-a)^{m+1}}{a+x}$$
we have: $$I=\int x^m\ln(a+x)dx$$ now with $$u=\ln(a+x)$$ we get: $$I=\int(e^u-a)^me^uu\,du$$ now using integration by parts: $$I=\frac{u(e^u-a)^{m+1}}{m+1}-\int\frac{(e^u-a)^{m+1}}{m+1}du$$ now try using binomial expansion. One way of doing it would be by writing: $$(e^u-a)^{m+1}=e^{(m+1)u}(1-ae^{-u})^{m+1}$$
With one step of by-parts integration (on $$x^m$$), you get rid of the logarithm and reduce to an incomplete Beta integral. https://en.wikipedia.org/wiki/Beta_function (check the fifth property and the incomplete function).
This indirectly proves that for general $$m$$ there is no closed-form expression.