How to evaluate this sum $\sum_{k=1}^{n}{n \choose k}(-2)^k\cdot \frac{1-2^{-k}}{k}$ I would like to work out this integral
$$\int_{0}^{\infty}\frac{1}{\left(x+\sqrt{1+x^2}\right)^{2n+1}}\large\frac{dx}{1+x^2}$$
Let start with $x=\tan(u)$, $u=\arctan(u)$
$dx=\sec^2(u)du$
$$\int_{0}^{\pi/2}\frac{\sec^2(u)}{(1+\tan^2(u))\left[\sqrt{1+\tan^2(u)}+\tan(u)\right]^{2n+1}}du$$
trig identity: $1+\tan^2(x)=\sec^2(x)$
$$\int_{0}^{\pi/2}\frac{du}{[\tan(u)+\sec(u)]^{2n+1}}$$
notice that $$\frac{1}{(\tan(u)+\sec(u))^{2n+1}}=\cos(u)\frac{(\sin(u)-1)^{n}}{(\sin(u)+1)^{n+1}}$$
$$\int_{0}^{\pi/2}\cos(u)\frac{(\sin u-1)^{n}}{(\cos u+1)^{n+1}}du$$
We do another sub: $v=\sin(u)$, $du=\frac{1}{\cos v} dv$
$$\int_{0}^{1}\frac{(v-1)^{n}}{(v+1)^{n+1}}dv$$
We still can further simplify this integral by sub: $y=v+1$, $dv=dy$
$$\int_{1}^{2}\frac{(y-2)^n}{y^{n+1}}dy$$
Applying binomial theorem
$$\frac{(y-2)^n}{y^{n+1}}=\sum_{k=0}^{n}{n \choose k}(-2)^ky^{-1-k}$$
$$\sum_{k=0}^{n}{n\choose k}(-2)^k{\int_{1}^{2}y^{-1-k}dy}=\sum_{k=1}^{n}{n\choose k}(-2)^k{\int_{1}^{2}y^{-1-k}dy}+\int_{1}^{2}y^{-1}dy$$
$$\sum_{k=1}^{n}{n \choose k}(-2)^k\cdot \frac{1-2^{-k}}{k}+\ln 2$$
I am stuck with this sum. Evaluating sum it is something I find it quite difficult.
I was checking on Harmonic number I got part of the answer:
$$\sum_{k=1}^{n}{n \choose k}(-2)^k\cdot \frac{1-2^{-k}}{k}+\ln 2=\sum_{k=1}^{n}{n \choose k}\frac{(-2)^k}{k}-\sum_{k=1}^{n}{n \choose k}\frac{(-1)^k}{k}+\ln 2$$
$$\sum_{k=1}^{n}{n \choose k}\frac{(-2)^k}{k}+H_n+\ln 2$$
 A: Through the substitution $x=\sinh t$ the original integral becomes
$$ \int_{0}^{+\infty}\frac{\cosh t}{1+\sinh^2 t}e^{-(2n+1) t}\,dt=\int_{0}^{+\infty}\frac{dt}{e^{(2n+1)t}\cosh t}\stackrel{t\mapsto \log u}{=}\int_{1}^{+\infty}\frac{2\,du}{u^{2n+1}(u^2+1)} $$
related to the Laplace transform of $\frac{1}{\cosh t}$. By letting $u\mapsto\frac{1}{v}$ we get
$$ \int_{0}^{1}\frac{2v}{v^2+1}\cdot v^{2n}\,dv = 2\int_{0}^{\pi/4}\left(\tan\theta\right)^{2n+1}\,d\theta=\int_{0}^{1}\frac{2v}{v^2+1}\left[(v^2+1)-1\right]^n\,dv$$
(so our function of $n$ is positive, decreasing and log-convex) or
$$ (-1)^n \log(2)+\sum_{k=1}^{n}\binom{n}{k}(-1)^{n-k}\int_{0}^{1}2v(v^2+1)^{k-1}\,dv $$
or
$$ (-1)^n \log(2)+(-1)^n \sum_{k=1}^{n}\binom{n}{k}(-1)^k \frac{2^k-1}{k} $$
which essentially is a binomial transform. Since $\frac{2^k-1}{k}=\int_{1}^{2}w^{k-1}\,dw$, the previous line equals
$$(-1)^n \log(2)+(-1)^n \int_{1}^{2}\frac{-1+(1-w)^n}{w}\,dw $$
or
$$ \frac{H_{n/2}-H_{(n-1)/2}}{2}=\frac{\psi((n+2)/2)-\psi((n+1)/2)}{2}=\sum_{m\geq 0}\frac{1}{(2m+n+1)(2m+n+2)} $$
or (assuming $n\in\mathbb{N}^+$)
$$ (-1)^n \sum_{k>n}\frac{(-1)^{k+1}}{k}. $$
A: Hint.
$$
(1-x)^n = \sum_{k=0}^{n}{n \choose k}(-1)^k x^k
$$
and
$$
\int (1-x)^n dx = \sum_{k=0}^{n}{n \choose k}(-1)^k \frac{x^{k+1}}{k+1}
$$
