Integral of (6x-1)dx/sqrt(x^2+3x+8) ?

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Integral of (6x-1)dx/sqrt(x^2+3x+8) ?

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Answer 1 · 2018-02-25T17:45:32

$6 \sqrt{x^{2} + 3 x + 8}$ $6sqrt(x^2+3x+8)$ - $\frac{10 \sqrt{23}}{23} ln (\sqrt{4 x^{2} + 12 x + 32} + 2 x + 3) + C$ $(10sqrt23)/23ln(sqrt(4x^2+12x+32)+2x+3)+C$

Explanation:

$\int \frac{6 x - 1}{\sqrt{x^{2} + 3 x + 8}} \cdot d x$ $int (6x-1)/sqrt(x^2+3x+8)*dx$

= $\int \frac{12 x - 2}{\sqrt{4 x^{2} + 12 x + 32}} \cdot d x$ $int (12x-2)/sqrt(4x^2+12x+32)*dx$

= $\int \frac{12 x - 2}{\sqrt{{(2 x + 3)}^{2} + 23}} \cdot d x$ $int (12x-2)/sqrt((2x+3)^2+23)*dx$

= $\int \frac{12 x + 18 - 20}{\sqrt{{(2 x + 3)}^{2} + 23}} \cdot d x$ $int (12x+18-20)/sqrt((2x+3)^2+23)*dx$

= $\frac{3}{2} \cdot \int \frac{8 x + 12}{\sqrt{{(2 x + 3)}^{2} + 23}} \cdot d x$ $3/2*int (8x+12)/sqrt((2x+3)^2+23)*dx$ - $\int \frac{20}{\sqrt{{(2 x + 3)}^{2} + 23}} \cdot d x$ $int 20/sqrt((2x+3)^2+23)*dx$

$A = \frac{3}{2} \cdot \int \frac{8 x + 12}{\sqrt{{(2 x + 3)}^{2} + 23}} \cdot d x$ $A=3/2*int (8x+12)/sqrt((2x+3)^2+23)*dx$

After using $y = 4 x^{2} + 12 x + 32$ $y=4x^2+12x+32$ and $d y = (8 x + 12) \cdot d x$ $dy=(8x+12)*dx$ transforms, $A$ $A$ becomes

$A = \frac{3}{2} \cdot \int \frac{d y}{\sqrt{y}}$ $A=3/2*int dy/sqrty$

= $3 \sqrt{y}$ $3sqrty$

= $3 \sqrt{4 x^{2} + 12 x + 32}$ $3sqrt(4x^2+12x+32)$

= $6 \sqrt{x^{2} + 3 x + 8}$ $6sqrt(x^2+3x+8)$

$B = \int \frac{20}{\sqrt{{(2 x + 3)}^{2} + 23}} \cdot d x$ $B=int 20/sqrt((2x+3)^2+23)*dx$

= $10 \int \frac{2}{\sqrt{{(2 x + 3)}^{2} + 23}} \cdot d x$ $10int 2/sqrt((2x+3)^2+23)*dx$

After using $2 x + 3 = \sqrt{23} \cdot tan z$ $2x+3=sqrt23*tanz$ and $2 d x = \sqrt{23} \cdot {(sec z)}^{2} \cdot d z$ $2dx=sqrt23*(secz)^2*dz$ transforms, $B$ $B$ becomes

$B = \int \frac{10 \sqrt{23} \cdot {(sec z)}^{2}}{23 sec z} \cdot d z$ $B=int (10sqrt23*(secz)^2)/(23secz)*dz$

= $\frac{10 \sqrt{23}}{23} \int sec z \cdot d z$ $(10sqrt23)/23int secz*dz$

= $\frac{10 \sqrt{23}}{23} \int \frac{sec z \cdot (sec z + tan z) \cdot d z}{sec z + tan z}$ $(10sqrt23)/23int (secz*(secz+tanz)*dz)/(secz+tanz)$

= $\frac{10 \sqrt{23}}{23} ln (sec z + tan z) - C 1$ $(10sqrt23)/23ln(secz+tanz)-C1$

After using $2 x + 3 = \sqrt{23} \cdot tan z$ $2x+3=sqrt23*tanz$ , $tan z = \frac{2 x + 3}{\sqrt{23}} and$ $tanz=(2x+3)/sqrt23 and$ secz= $\frac{\sqrt{4 x^{2} + 12 x + 32}}{\sqrt{23}}$ $sqrt(4x^2+12x+32)/sqrt23$ inverse transforms,

$B = \frac{10 \sqrt{23}}{23} ln (\sqrt{4 x^{2} + 12 x + 32} + 2 x + 3) - C$ $B=(10sqrt23)/23ln(sqrt(4x^2+12x+32)+2x+3)-C$

Thus,

$\int \frac{6 x - 1}{\sqrt{x^{2} + 3 x + 8}} \cdot d x$ $int (6x-1)/sqrt(x^2+3x+8)*dx$

= $A - B$ $A-B$

= $6 \sqrt{x^{2} + 3 x + 8}$ $6sqrt(x^2+3x+8)$ - $\frac{10 \sqrt{23}}{23} ln (\sqrt{4 x^{2} + 12 x + 32} + 2 x + 3) + C$ $(10sqrt23)/23ln(sqrt(4x^2+12x+32)+2x+3)+C$

Note: $C = C 1 + ln (\sqrt{23})$ $C=C1+Ln(sqrt23)$

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Integral of (6x-1)dx/sqrt(x^2+3x+8) ?

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