Evaluate the integral: I = integral

Evaluate the integral: $$ I = \int \frac1{1-2x}\sqrt{\frac{1+2x}{1-2x}}\,dx = \int\frac{\sqrt{1+2x}}{(1-2x)^{3/2}}\,dx. $$

Step 1 · Substitute

Let $2x=\sin\varphi$, so $dx=\tfrac12\cos\varphi\,d\varphi$, $1+2x=1+\sin\varphi$, $1-2x=1-\sin\varphi$:

$$ I = \frac12\int\frac{\sqrt{1+\sin\varphi}\,\cos\varphi}{(1-\sin\varphi)^{3/2}}\,d\varphi. $$

Step 2 · Simplify

Since $\cos\varphi=\sqrt{(1-\sin\varphi)(1+\sin\varphi)}$ (for $-\tfrac\pi2<\varphi<\tfrac\pi2$):

$$ \frac{\sqrt{1+\sin\varphi}\,\cos\varphi}{(1-\sin\varphi)^{3/2}} = \frac{(1+\sin\varphi)\sqrt{1-\sin\varphi}}{(1-\sin\varphi)^{3/2}} = \frac{1+\sin\varphi}{1-\sin\varphi}. $$

Step 3 · Expand and integrate

Multiplying numerator and denominator by $(1+\sin\varphi)$: $\dfrac{1+\sin\varphi}{1-\sin\varphi}=\dfrac{(1+\sin\varphi)^2}{\cos^2\varphi}=2\sec^2\varphi+2\sec\varphi\tan\varphi-1$.

$$ I = \frac12\int\left(2\sec^2\varphi+2\sec\varphi\tan\varphi-1\right)d\varphi = \tan\varphi+\sec\varphi-\frac\varphi2+C. $$

Step 4 · Convert back to $x$

Since $\sin\varphi=2x$, $\cos\varphi=\sqrt{1-4x^2}$, $\tan\varphi=2x/\sqrt{1-4x^2}$, $\sec\varphi=1/\sqrt{1-4x^2}$, $\varphi=\arcsin(2x)$:

$$ I = \frac{1+2x}{\sqrt{1-4x^2}} - \frac12\arcsin(2x) + C. $$

Summary Table

Additional Notes

• Major rewrite: the source's chain of substitutions ($u=1-\sin\theta$, then $v=\sqrt u$) led to a final answer of $\sqrt{(1+2x)/(1-2x)}-\arcsin\sqrt{(1-2x)/2}$. Differentiating that at $x=0$ gives $1-1=0$, while the true integrand at $x=0$ is $1$ — confirming the error. The direct trigonometric approach above, using $2x=\sin\varphi$ from the start, avoids the intermediate substitution errors.
• The domain requires $-\tfrac12\leq x<\tfrac12$.
• The result was verified by differentiation: combining the derivatives of $(1+2x)/\sqrt{1-4x^2}$ and $-\tfrac12\arcsin(2x)$ over the common denominator $(1-4x^2)^{3/2}$ gives numerator $(1+2x)^2$, and $(1+2x)^2/(1-4x^2)^{3/2}=\sqrt{1+2x}/(1-2x)^{3/2}$, matching the integrand exactly.