Evaluation of $\int\frac{1}{1+(x+1)^{{1}/{n}}}dx$ for $n\in \mathbb{N},$

Question

Evaluate $$\int\frac{1}{1+(x+1)^{{1}/{n}}}\,\mathrm dx$$ for $n\in \mathbb{N}$

$\bf{My\; Try::}$ Let $$(x+1)=t^n\;,$$ Then $$dx = nt^{n-1}dt$$

So $$\displaystyle I = n\int\frac{t^{n-1}}{1+t}dt\;,$$ Now $$1+t=y\;,$$ Then $$dt = dy$$

So $$\displaystyle I = n\int\frac{(y-1)^{n-1}}{y}dy$$

Now How can i solve after that, Help me, Thanks

Answer 1

$$I=\int\frac{1}{1+(x+1)^{\frac{1}{n}}}dx$$ Let $t^n=x+1;nt^{n-1}dt=dx$ $$I=n\int\frac{t^{n-1}}{1+t}dt$$ Let $u=t+1;du=dt$ $$I=n\int\frac{(u-1)^{n-1}}{u}du=n\int\sum_{k=0}^{n-1}\binom{n-1}{k}u^{k-1}(-1)^{n-1-k}du=n\sum_{k=1}^{n-1}\binom{n-1}{k}\frac{u^{k}}{k}(-1)^{n-1-k}+\binom{n-1}0\ln|u|(-1)^{n-1}+c$$

Answer 2

Not a full answer, but you can reduce this to a trigonometric integral fairly simply by supposing $(x+1)^{1/n}=\tan^2\theta$. With the simplification, you obtain $I=2n\int\tan^{n-1}\theta\,d\theta$. As you can see, the answer is not very pretty: WolframAlpha output

Answer 3

Use the geometric series:

$$\frac1{1-r}=1+r+r^2+r^3+\dots$$

$$\frac1{1+t}=1-t+t^2-t^3+\dots$$

$$\frac1{1+(x+1)^{1/n}}=1-(x+1)^{1/n}+(x+1)^{2/n}-\dots=\sum_{k=0}^\infty(-1)^k(x+1)^{k/n}$$

$$\int\frac1{1+(x+1)^{1/n}}dx=\int\sum_{k=0}^\infty(-1)^k(x+1)^{k/n}dx$$

$$=\sum_{k=0}^\infty(-1)^k\int(x+1)^{k/n}dx$$

$$=c+\sum_{k=0}^\infty(-1)^k\frac{n(x+1)^{(k+n)/n}}{k+n}$$