Find the value of $\int \frac{x^{n}\operatorname{ln}(x)}{(x^{n+1}+1)^{n}}dx$, where $n $ is any natural number different from $0,1,2$

Question

I had to find the value of $\int \frac{x^{n}\operatorname{ln}(x)}{(x^{n+1}+1)^{n}}dx$, where $n $ is any natural number different from $0,1,2$ and $ x$ is a positive, real number. By a change of variable from $x$ to $t^{\frac{1}{n+1}}$, we obtain $\frac{1}{(n+1)^2}\int \frac{\operatorname{ln}(t)}{(t+1)^n}dt$. From here I get, integrating by parts, a form containing $\int \frac{1}{t(t+1)^{n-1}}dt$, to which I got stuck.

Answer 1

Note that for $j\ge 3$ $$\begin{align*} \int \frac1{t(1+t)^{j-1}}\mathrm dt&=\int \frac1{t(1+t)^{j-2}}-\frac1{(1+t)^{j-1}}\mathrm dt \\&=\int\frac1{t(1+t)^{j-2}}\mathrm dt+\frac1{j-2}\frac1{(1+t)^{j-2}}. \end{align*}$$ By telescoping, we can compute $\int \frac1{t(1+t)^{n-1}}\mathrm dt$ as follows: $$\begin{align*} \int \frac1{t(1+t)^{n-1}}\mathrm dt&=\int \frac 1 {t(1+t)} \mathrm dt+\sum_{j=3}^n\int \frac {\mathrm dt}{t(1+t)^{j-1}}-\int \frac {\mathrm dt}{t(1+t)^{j-2}} \\&=\int \frac 1 t-\frac1{t+1} \mathrm dt+\sum_{j=3}^n\frac 1 {j-2}\frac1{(1+t)^{j-2}} \\&=\ln\left(\frac t {1+t}\right)+\sum_{j=1}^{n-2}\frac 1 {j}\frac1{(1+t)^j}. \end{align*}$$