Evaluate the integral: I = integral (arcsin

Evaluate the integral: $$ I = \int \frac{\arcsin x}{\sqrt{1+x}}\,dx. $$

Step 1 · Set up integration by parts

Let:

$$ u = \arcsin x \quad \Rightarrow \quad du = \frac{dx}{\sqrt{1-x^2}}, \qquad dv = \frac{dx}{\sqrt{1+x}} \quad \Rightarrow \quad v = 2\sqrt{1+x}. $$

Applying the formula:

$$ I = 2\sqrt{1+x}\arcsin x - \int \frac{2\sqrt{1+x}}{\sqrt{1-x^2}}\,dx. $$

Step 2 · Simplify the leftover integral

Since $\sqrt{1-x^2}=\sqrt{1-x}\sqrt{1+x}$, the $\sqrt{1+x}$ factors cancel:

$$ \frac{2\sqrt{1+x}}{\sqrt{1-x^2}} = \frac{2\sqrt{1+x}}{\sqrt{1-x}\sqrt{1+x}} = \frac{2}{\sqrt{1-x}}. $$

Step 3 · Integrate the remainder

Let $w=1-x$, $dw=-dx$:

$$ \int \frac{2}{\sqrt{1-x}}\,dx = -2\int w^{-1/2}\,dw = -4\sqrt w = -4\sqrt{1-x}. $$

Step 4 · Combine

$$ I = 2\sqrt{1+x}\arcsin x - \left(-4\sqrt{1-x}\right) + C = 2\sqrt{1+x}\arcsin x + 4\sqrt{1-x} + C. $$

Summary Table

Additional Notes

• The domain requires $-1<x\leq1$, where both $\arcsin x$ and $1/\sqrt{1+x}$ are defined.
• The key simplification is recognizing $1-x^2=(1-x)(1+x)$ under the square root, which lets the awkward $\sqrt{1+x}$ factor introduced by integration by parts cancel against the $\sqrt{1+x}$ hiding inside $\sqrt{1-x^2}$, leaving an elementary $1/\sqrt{1-x}$ integral instead of a genuinely new trigonometric substitution.
• The result can be checked by differentiating $2\sqrt{1+x}\arcsin x+4\sqrt{1-x}$: this gives $\dfrac{\arcsin x}{\sqrt{1+x}}+\dfrac{2}{\sqrt{1-x}}-\dfrac{2}{\sqrt{1-x}}=\dfrac{\arcsin x}{\sqrt{1+x}}$, matching the integrand.