∫6x+56+x−2x2dx=Put 6x+5=Addx6+x−2x2+8⇒6x+5=A(−4x+1)+B⇒−4A=6,A+B=5⇒A=−32,B=5+32=132. ∴6x+5=−32ddx6+x−2x2+132∫6x+56+x−2x2dx=−32∫ddx6+x−2x26+x−2x2+132∫16+x−2x2dx=−32⋅26+x−2x2+132∫1−2x2−x2−3dx=−36+x−2x2+132⋅12∫1−x−142−3−116dx=−36+x−2x2+1322∫1742−x−142dx=−36+x−2x2+1322Sin−1⁡x−1/47/4+c=−36+x−2x2+1322Sin−1⁡4x−17+c.

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$\int \frac{6 x + 5}{\sqrt{6 + x - 2 x^{2}}} d x =$

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$\begin{aligned} - 3 \sqrt{6 + x - 2 x^{2}} + \frac{13}{2 \sqrt{2}} {Sin}^{- 1} (\frac{4 x - 1}{7}) + c \end{aligned}$

$\begin{aligned} - 3 \sqrt{6 + x - 2 x^{2}} + \frac{13}{\sqrt{2}} {Sinh}^{- 1} (\frac{4 x - 1}{7}) + c \end{aligned}$

$- 3 \sqrt{6 + x - 2 x^{2}} + \frac{13}{2 \sqrt{3}} {Sinh}^{- 1} (\frac{4 x + 1}{7}) + c$

$3 \sqrt{6 + x - 2 x^{2}} - \frac{13}{2 \sqrt{2}} {Cos}^{- 1} (\frac{4 x - 1}{7}) + c$

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Detailed Solution

$\int \frac{6 x + 5}{\sqrt{6 + x - 2 x^{2}}} d x =$

Put $6 x + 5 = A \frac{d}{d x} (6 + x - 2 x^{2}) + 8 \Rightarrow 6 x + 5 = A (- 4 x + 1) + B \Rightarrow - 4 A = 6, A + B = 5$

$\begin{array}{l} \Rightarrow A = - \frac{3}{2}, B = 5 + \frac{3}{2} = \frac{13}{2} . ∴ 6 x + 5 = - \frac{3}{2} \frac{d}{d x} (6 + x - 2 x^{2}) + \frac{13}{2} \\ \int \frac{6 x + 5}{\sqrt{6 + x - 2 x^{2}}} d x = - \frac{3}{2} \int \frac{\frac{d}{d x} (6 + x - 2 x^{2})}{\sqrt{6 + x - 2 x^{2}}} + \frac{13}{2} \int \frac{1}{\sqrt{6 + x - 2 x^{2}}} d x \\ = - \frac{3}{2} \cdot 2 \sqrt{6 + x - 2 x^{2}} + \frac{13}{2} \int \frac{1}{\sqrt{- 2 (x^{2} - \frac{x}{2} - 3)}} d x \\ = - 3 \sqrt{6 + x - 2 x^{2}} + \frac{13}{2} \cdot \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{- [{(x - \frac{1}{4})}^{2} - 3 - \frac{1}{16}]}} d x \end{array}$

$\begin{array}{l} = - 3 \sqrt{6 + x - 2 x^{2}} + \frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{{(\frac{7}{4})}^{2} - {(x - \frac{1}{4})}^{2}}} d x = - 3 \sqrt{6 + x - 2 x^{2}} + \frac{13}{2 \sqrt{2}} {Sin}^{- 1} (\frac{x - 1 / 4}{7 / 4}) + c \\ = - 3 \sqrt{6 + x - 2 x^{2}} + \frac{13}{2 \sqrt{2}} {Sin}^{- 1} (\frac{4 x - 1}{7}) + c . \end{array}$