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Pattern - 42 of Indefinite Integration

February 24, 2021
Category: Pattern - 42 of Indefinite Integration

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⋆ Pattern – 42

$\int\frac{ax^2+bx+c}{\left(qx+4\right)\sqrt{px^2+1x+r}}\ dx$

Substitute :

Nr = A(gx + h) $\frac{d}{dx}$ ( p $x^2$ + qx + r ) + B (gx + h) + c

1) Evaluate : $\ \int\frac{2x^2+7x+11}{\left(x+2\right)\ \sqrt{x^2+4x+8}}\ dx$

Solution :

I= $\ \int\frac{2x^2+7x+11}{\left(x+2\right)\ \sqrt{x^2+4x+8}}\ dx$

Let, 2 $x^2$ + 7x + 11 = A(x+2) (2x+4) + B(x+2) + C

∴ 2 $x^2$ + 7x + 11 = A (2 $x^2$ + 8x + 8 ) + Bx + 2B + C

= 2A $x^2$ + (8A+B) x +(8A + 2B + C)

∴ 2A = 2

∴ A = 1

8A + B = 7

∴ B = -1

8A + 2B + C = 11

∴ 8 – 2 + C = 11

∴ c = 5

$\therefore\ \ I=\ \int\frac{\left(x+2\right)\left(2x+4\right)-\ \left(x+2\right)+5}{\left(x+2\right)\sqrt{x^2+4x+8}}\ dx$

$=\ \int\frac{1x+4}{\sqrt{x^2+4x+8}}dx$ $-\ \int\frac{1}{\sqrt{x^2+4x+8}}dx$ $+5\int\frac{1}{\left(x+2\right)\sqrt{x^2+4x+8}}\ dx$

$=\ \int{\left(x^2+4x+8\right)^\frac{-1}{2}\ \left(2x+4\right)}\ dx-\ \int\frac{1}{\sqrt{\left(x+2\right)^2+4}}dx+5I_1--\left(1\right)$

Where,

$I_1=\ \int\frac{1}{\left(x+2\right)\ \sqrt{x^2+4x+8}}\ dx$

x+2= $\ \frac{1}{t}$

$\therefore\ \ dx=\ \frac{-1}{t^2}\ dt$

$=\ \frac{1-2t}{t}$

$\sqrt{x^2+4x+8}=\ \sqrt{\left(\frac{1-2t}{t}\right)^2+4\left(\frac{1-2t}{t}\right)+8}$

$=\ \sqrt{\frac{1-4t+4t^2}{t^2}+\ \frac{4-8t}{t}+8}$

$=\ \sqrt{\frac{1-4t+4t^2+4t-5t^2+8t^2}{t^2}}$

$=\ \frac{\sqrt{4t^2+1}}{t}$

$\therefore\ \ I_1=\ -\int\frac{t\times t}{\sqrt{4t^2+1}}\times\frac{1}{t^2}dt$

$=\ -\ \int\frac{1}{\sqrt{\left(2t\right)^2+\ \left(1\right)^2}}\ dt$

$=\ \frac{-1}{2}\log{\left|2t+\ \sqrt{4t^2+1}\right|}+c$

Here ,

I = $\ \int{\left(x^2+4x+8\right)^\frac{-1}{2}\ \left(2x+4\right)}\ dx-\ \int\frac{1}{\sqrt{\left(x+2\right)^2+\ \left(2\right)^2}}\ dx+5I_1$

————(1)

Here,

$I_1=\ \frac{-1}{2}\log{\left|2t+\ \sqrt{4t^2+1}\right|}$

$=\ \frac{1}{2}\log{\left|\frac{2}{x+2}+\ \sqrt{\frac{4}{\left(x+2\right)^2}+1}\right|}$

$=\ \frac{1}{2}\log{\left|\frac{2}{x+2}+\ \sqrt{\frac{4}{\left(x+2\right)^2}+\frac{4}{4}}\right|}$

$=\ \frac{1}{2}\log{\left[2\left|\frac{1}{x+2}+\ \sqrt{\frac{1}{\left(x+2\right)^2}+\frac{1}{4}}\right|\right]}$

$=\frac{1}{2}\log{2}+\frac{1}{2}\log{\left|\frac{1}{x+2}+\ \sqrt{\frac{1}{\left(x+2\right)^2}+\frac{1}{4}}\right|}$

$I= \frac{\left(x^2+4x+8\right)^\frac{1}{2}}{\frac{1}{2}}$ $-\log{\left|\left(x+2\right)+\sqrt{x^2+4x+8}\right|}+$ $\frac{5}{2}\log{\left|\frac{1}{x+2}+\ \sqrt{\frac{1}{\left(x+2\right)^2}+\ \frac{1}{4}}\right|}+c$

$=2\sqrt{x^2+4x+8}-\log{\left|\left(x+2\right)+\ \sqrt{x^2+4x+8}\right|}$ + $\ \frac{5}{2}\log{\left|\frac{1}{x+2\ }+\ \sqrt{\frac{1}{\left(x+2\right)^2}+\frac{1}{4}}\right|}+c$

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Pattern - 42 of Indefinite Integration

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