Q.$\displaystyle\int_0^{1/\sqrt2} \dfrac{e^{\sin^{-1}x}\sin^{-1}x}{\sqrt{1-x^2}}\,dx$

Bernoulli's formula extends integration by parts to integrate $\int u(x)v(x)\,dx$ in one pass, whenever $u(x)$ is a polynomial (so its successive derivatives eventually vanish) and $v(x)$ can be integrated repeatedly with ease (e.g. $\sin nx,\ \cos nx,\ e^{ax}$).

Notation. Write $u^{(1)}=\dfrac{du}{dx},\ u^{(2)}=\dfrac{du^{(1)}}{dx},\ u^{(3)}=\dfrac{du^{(2)}}{dx},\ldots$ for the successive derivatives of $u$, and $v_{(1)}=\int v\,dx,\ v_{(2)}=\int v_{(1)}\,dx,\ v_{(3)}=\int v_{(2)}\,dx,\ldots$ for the successive anti-derivatives of $v$.

Bernoulli's formula.

— an alternating sum that terminates because $u$ is a polynomial: once a derivative $u^{(m)}$ hits $0$, every later term vanishes.

Derivation sketch. Apply integration by parts once with $dv_{(1)}=v\,dx$ to get $\int uv\,dx=uv_{(1)}-\int u^{(1)}v_{(1)}\,dx$; then apply by-parts again to $\int u^{(1)}v_{(1)}\,dx$ with $dv_{(2)}=v_{(1)}\,dx$, and so on — each step trades one derivative of $u$ for one further anti-derivative of $v$. …