Approximating the integral $\int_1^\frac{3}{2}\frac{\ln(x+1)-\ln(2)}{x-1}dx$ to accuracy of $0.001$ I need to approximate the integral $\int\limits_1^\frac{3}{2}\frac{\ln(x+1)-\ln(2)}{x-1}dx$ with an accuracy of $0.001.$
How don't know how I can do it, since the integrand doesn't have an elementary antiderivative, and because it's an improper integral.
I thought about trying to use Taylor Series, but I can't expand the $\ln(1+x)$ because $|x|>1.$
How can I do it?
 A: $$I=\int_{1}^{3/2} \frac{\ln(1+x)-\ln 2}{x-1}dx= \int_{0}^{1/2} \frac{\ln (1+u/2)}{u} du.$$
We have used $u=x-1$. Next use $\ln(1+z) \approx z-z^2/2+z^3/3$ for an accuracy of about $0.01$, then
$$I_{approx}=\int_{0}^{1/2} [1/2-u/8+u^2/(24)] du=[u/2-u^2/(16)+u^3/(72)]_{u=0}^{1/2}=\frac{17}{72}=0.2361$$
Whereas its exact value as per Mathematica is $0.2359$.
A: $$
\begin{align}
\int_1^{3/2}\frac{\log(x+1)-\log(2)}{x-1}\,\mathrm{d}x
&=\int_0^{1/2}\frac{\log(x+2)-\log(2)}x\,\mathrm{d}x\tag1\\
&=\int_0^{1/4}\frac{\log(x+1)}x\,\mathrm{d}x\tag2\\
&=\int_0^{1/4}\sum_{k=1}^\infty(-1)^{k-1}\frac{x^{k-1}}k\,\mathrm{d}x\tag3\\
&=\sum_{k=1}^\infty(-1)^{k-1}\frac{(1/4)^k}{k^2}\tag4
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x+1$
$(2)$: substitute $x\mapsto2x$
$(3)$: use the Taylor series for $\frac{\log(1+x)}x$
$(4)$: integrate term by term
For $k=4$, $(-1)^{k-1}\frac{(1/4)^k}{k^2}=-\frac1{4096}$, so the Alternating Series Test says we only need to include the first three terms:
$$
\begin{align}
\sum_{k=1}^3(-1)^{k-1}\frac{(1/4)^k}{k^2}
&=\frac{17}{72}\tag5\\
&\approx0.236111\tag6
\end{align}
$$
Adding more terms, we get that
$$
\begin{align}
\sum_{k=1}^\infty(-1)^{k-1}\frac{(1/4)^k}{k^2}
&=-\operatorname{Li}_2(-1/4)\tag7\\
&\approx0.235900297686\tag8
\end{align}
$$
Thus, the error is $0.000211\le\frac1{4096}$ as the Alternating Series Test promised.
A: To your Taylor series idea:  Since $x \in [1,3/2]$, $x+1 \in [2,5/2]$  But this means $\frac{1}{x+1} \in [2/5, 1/2]$ which is in the radius of convergence of the usual Taylor series of the natural logarithm (centered at $1$).  So use
$$  \ln(x+1) = - \ln \frac{1}{x+1}  $$
to rewrite your integrand to use your Taylor series.
Alternatively, you could construct the Taylor series centered at, say, $2$:
$$  \ln 2 + \frac{x-2}{2} - \frac{(x-2)^2}{8} + \frac{(x-2)^3}{24} - {\dots}  \text{.}  $$  Those numerators are upper bounded by powers of $1/2$ for your interval or integration and the denominators are approximately doubling at each step, so you shouldn't need more than about six terms...
Regarding having an elementary antiderivative:  The antiderivative of the integrand is $$\ln(1/2 - x/2) \ln (1/2+x/2) \mathrm{Li}_2(x/2+1/2)  \text{,}  $$
where $\mathrm{Li}_2$ is the dilogarithm.  There are many techniques for evaluating the dilogarithm and the link provides a few starting points you could use.
Another method is to approximate with a (finite) Riemann sum.  It is elementary calculus (optimization) to show that the the derivative of the integrand has bounded magnitude.  (Just examining the graph of the derivative of the integrand, the magnitude is bounded by $0.13$ and this visual estimate can be improved by using actual calculation).  Then an error bound on the evenly-spaced $n$ term Riemann sum approximation to the integral, is $\frac{1}{2n}(0.13)$.  This is less than $0.001$ when $n > 65$, so look at (the right Riemann sum)
$$  \sum_{n=1}^{65} f \left( 1 + \frac{n}{65}(3/2 - 1)\right) \cdot \frac{3/2 - 1}{65}  \text{,}  $$
where $f(x)$ is your integrand.  The error in using this series is actually $0.000\,206{\dots}$.
A: Developed as an infinite series around $x=1$
$$\frac{\ln(x+1)-\ln(2)}{x-1}=\sum_{n=0}^\infty\frac{(-1)^n }{2^{n+1}(n+1)}(x-1)^n$$
$$I=\int_1^\frac{3}{2}\frac{\ln(x+1)-\ln(2)}{x-1}dx=\sum_{n=0}^\infty\frac{(-1)^n }{2^{2 (n+1)}\,(n+1)^2}$$ So, if you want an accuracy of $\epsilon$, write
$$I=\sum_{n=0}^p\frac{(-1)^n }{2^{2 (n+1)}\,(n+1)^2}+\sum_{n=p+1}^\infty\frac{(-1)^n }{2^{2 (n+1)}\,(n+1)^2}$$ and since this is an alternating series, you search for $p$ such that
$$R_p=\frac{1}{2^{2 (p+2)}\,(p+2)^2} \leq \epsilon$$ that is to say
$$p=-2+\frac{W\left(t\right)}{\log (2)} \qquad \text{where} \qquad t=\frac{\log (2)}{\sqrt{\epsilon }}$$ where appears Lambert function and, as usual, you will need to use $\lceil p \rceil$.
For the estimation of $W(t)$, we can use the expansion given in the Wikipedia page
$$W(t)=L_1-L_2+\frac{L_2}{L_1}+\frac{(L_2-2) L_2}{2 L_1^2}+\frac{(2 L_2^2-9L_2+6) L_2}{6 L_1^3}+ ...$$ where $L_1=\log(t)$ and $L_2=\log(L_1)$.
If $\epsilon=0.001$, this will give $p=1.2725$ and then $p=2$. Checking the various values of $R_p$
$$R_1=\frac 1 {576} \qquad R_2=\frac 1 {4096} \qquad R_3=\frac 1 {25600} \qquad R_4=\frac 1 {147456}$$
So, using $p=2$, the approximate value of the integral is
$$\frac{1}{4}-\frac{1}{64}+\frac{1}{576}=\frac{17}{72}\approx 0.236111$$ while its exact value is
$$\frac{1}{6} \left(-6 \text{Li}_2\left(\frac{4}{5}\right)+\pi ^2+3 \log ^2(4)-3 \log
   ^2(5)\right)\approx 0.235900$$.
