Q.Evaluate: $\displaystyle\int \dfrac{6x+5}{\sqrt{1-4x-4x^2}}\, dx$ OR If $y = \sin^{-1}\dfrac{1}{2}\left(\sqrt{1+x}+\sqrt{1-x}\right)$ then show that $\dfrac{dy}{dx} = \dfrac{-1}{2\sqrt{1-x^2}}$.

Writing $6x+5$ as a combination of $\dfrac{d}{dx}(1-4x-4x^2)=-4-8x$ and a constant splits the integral into a square-root term and an arcsine term.

Let $u=1-4x-4x^2$, so $\dfrac{du}{dx}=-4-8x$.

Write $6x+5 = A(-8x-4)+B$. Matching the $x$ coefficient: $-8A=6\implies A=-\dfrac34$. Matching the constant: $-4A+B=5\implies -4(-\tfrac34)+B=5\implies 3+B=5\implies B=2$.

So:

$\int\frac{6x+5}{\sqrt{1-4x-4x^2}}dx = -\frac34\int\frac{-8x-4}{\sqrt{1-4x-4x^2}}dx + 2\int\frac{dx}{\sqrt{1-4x-4x^2}}$

First piece: this is $-\dfrac34\int\dfrac{du}{\sqrt u} = -\dfrac34\cdot2\sqrt u = -\dfrac32\sqrt{1-4x-4x^2}$.

Second piece: complete the square: $1-4x-4x^2 = 2-4(x+\tfrac12)^2 = \left(\sqrt2\right)^2-\left(2(x+\tfrac12)\right)^2\cdot\frac12$... more directly, $1-4x-4x^2 = -4\left(x+\tfrac12\right)^2+2$, so:

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