computing an integral when the function is not defined at some point. In computing the integral $I=\displaystyle\int_0^1\arcsin(x)dx$ we are led to compute the integral $J=\displaystyle\int_0^1 \dfrac{x}{\sqrt{1-x^2}}dx$, while the initial integral $I$ is well defined because the function $arcsin(x)$ is continuous on the interval $[0,1]$, I have trouble working with the second integral $J$ because the function $f(x)=\dfrac{x}{\sqrt{1-x^2}}$ is not even defined at $x=1$ and can not be extended by continuity at $x=1$ because the limit of $f(x)$ as $x$ approaches $1$ is infinity. On the other hand I know that the function $f(x)$ has an antiderivative $F(x)=-\sqrt{1-x^2}$ so the integral $J$ is computable via the fundamental theorem of integration which gives that $J=1$. Thank you for explaining this problem !
 A: This is an improper integral and can be computed as:
\begin{equation}\begin{aligned}
J&= \lim_{b\to1} \int^b_0 {x \over \sqrt{1-x^2}}\ dx\\
 &= -{1\over 2} \lim_{b\to1} \int^{1-b}_1 u^{-1/2}\ du \\
 &= -{2\over 2} \lim_{b\to1} \bigg[{u^{1/2}}\bigg]^{1-b}_{1}\\
 &= -{2\over 2}(-1) \\
 &= 1\ .
\end{aligned}\end{equation}
Where the substitution $u=1-x^2$ is used for the second equality.
A: When you use integration by parts (Riemann version) you have to make sure everything appears in the integrals is good, in other words, the integrands have no singularities. A trick to avoid your concern is to consider the smaller intervals which the singularity does not fall into, and then apply Monotone Convergent Theorem:
\begin{align*}
\int_{0}^{1-1/n}\arcsin xdx&=x\arcsin x\bigg|_{x=0}^{x=1-1/n}-\int_{0}^{1-1/n}\frac{x}{\sqrt{1-x^{2}}}dx\\
&=\left(1-\frac{1}{n}\right)\arcsin\left(1-\frac{1}{n}\right)+\left(\sqrt{1-\left(1-\frac{1}{n}\right)^{2}}-1\right),
\end{align*}
and then we take $n\rightarrow\infty$. 
