e^x(x^-1x√1-x^2+^-1x(1-x^2)^3/2+(x)/(1-x^2))dx = g(x)+C.

Question

$\displaystyle\int e^x\left(\frac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac{\sin^{-1}x}{(1-x^2)^{3/2}}+\frac{x}{1-x^2}\right)dx = g(x)+C$. Then $g\left(\dfrac12\right)$ equals:

Accepted Answer

Since $\dfrac{d}{dx}\left(\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}\right) = \dfrac{\sin^{-1}x}{(1-x^2)^{3/2}}+\dfrac{x}{1-x^2}$, the integrand is of the form $e^x[h(x)+h'(x)]$ with $h(x)=\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}$, so

$$g(x) = e^x\cdot\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$$

(the constant is absorbed, taking $g(x)$ as this particular antiderivative to match the given answer choices).

$$g\left(\frac12\right) = e^{1/2}\cdot\frac{\frac12\cdot\frac\pi6}{\sqrt{3}/2} = \frac{\pi}{6}\sqrt{\frac{e}{3}}$$

Solutions · 1