How do you integrate $ln (x^{2} + 13 x + 40) d x$ $ln(x^2 + 13x + 40) dx$ ?

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How do you integrate $ln (x^{2} + 13 x + 40) d x$ $ln(x^2 + 13x + 40) dx$ ?

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Answer 1 · 2018-03-15T15:16:20

$\frac{2 x + 13}{2} \cdot ln (x^{2} + 13 x + 40) - 2 x + \frac{3}{2} ln (\frac{x + 8}{x + 5}) + C$ $(2x+13)/2*Ln(x^2+13x+40)-2x+3/2Ln((x+8)/(x+5))+C$

Explanation:

$\int ln (x^{2} + 13 x + 40) \cdot d x$ $int Ln(x^2+13x+40)*dx$

= $x ln (x^{2} + 13 x + 40) - \int x \cdot \frac{(2 x + 13) \cdot d x}{x^{2} + 13 x + 40}$ $xLn(x^2+13x+40)-int x*((2x+13)*dx)/(x^2+13x+40)$

= $x ln (x^{2} + 13 x + 40) - \int \frac{(2 x^{2} + 13 x) \cdot d x}{x^{2} + 13 x + 40}$ $xLn(x^2+13x+40)-int ((2x^2+13x)*dx)/(x^2+13x+40)$

= $x ln (x^{2} + 13 x + 40) - \int \frac{(2 x^{2} + 26 x + 80 - 13 x - 80) \cdot d x}{x^{2} + 13 x + 40}$ $xLn(x^2+13x+40)-int ((2x^2+26x+80-13x-80)*dx)/(x^2+13x+40)$

= $x ln (x^{2} + 13 x + 40) - \int 2 d x + \int \frac{(13 x + 80) \cdot d x}{x^{2} + 13 x + 40}$ $xLn(x^2+13x+40)-int 2dx+int ((13x+80)*dx)/(x^2+13x+40)$

= $x ln (x^{2} + 13 x + 40) - 2 x + C 1 + \int \frac{(52 x + 320) \cdot d x}{4 x^{2} + 52 x + 160}$ $xLn(x^2+13x+40)-2x+C1+int ((52x+320)*dx)/(4x^2+52x+160)$

= $x ln (x^{2} + 13 x + 40) - 2 x + C 1 + \int \frac{(26 x + 160) \cdot 2 d x}{{(2 x + 13)}^{2} - 3^{2}}$ $xLn(x^2+13x+40)-2x+C1+int ((26x+160)*2dx)/((2x+13)^2-3^2)$

$A = \int \frac{(26 x + 160) \cdot 2 d x}{{(2 x + 13)}^{2} - 3^{2}}$ $A=int ((26x+160)*2dx)/((2x+13)^2-3^2)$

After using $2 x + 13 = 3 sec y$ $2x+13=3secy$ , $2 d x = 3 sec y \cdot tan y \cdot d y$ $2dx=3secy*tany*dy$ and $x = \frac{3 sec y - 13}{2}$ $x=(3secy-13)/2$ transforms, $A$ $A$ became

$A = \int \frac{(26 \cdot \frac{3 sec y - 13}{2} + 160) \cdot 3 sec y \cdot tan y \cdot d y}{9 {(tan y)}^{2}}$ $A=int ((26*(3secy-13)/2+160)*3secy*tany*dy)/(9(tany)^2)$

= $\int \frac{(39 sec y - 9) \cdot sec y \cdot d y}{3 tan y}$ $int ((39secy-9)*secy*dy)/(3tany)$

= $13 \int \frac{{(sec y)}^{2} \cdot d y}{tan y} - 3 \int \frac{sec y \cdot d y}{tan y}$ $13int((secy)^2*dy)/(tany)-3int (secy*dy)/tany$

= $13 ln (tan y) - 3 \int csc y \cdot d y$ $13Ln(tany)-3int cscy*dy$

= $\frac{13}{2} ln ({(tan y)}^{2}) - 3 \int \frac{csc y \cdot (csc y + cot y) \cdot d y}{csc y + cot y}$ $13/2Ln((tany)^2)-3int (cscy*(cscy+coty)*dy)/(cscy+coty)$

= $\frac{13}{2} ln ({(sec y)}^{2} - 1) + 3 ln (csc y + cot y)$ $13/2Ln((secy)^2-1)+3Ln(cscy+coty)$

= $\frac{13}{2} ln ({(sec y)}^{2} - 1) + 3 ln (\frac{sec y + 1}{tan y})$ $13/2Ln((secy)^2-1)+3Ln((secy+1)/tany)$

= $\frac{13}{2} ln ({(sec y)}^{2} - 1) + 3 ln ⎛ ⎜ ⎜ ⎝ \frac{sec y + 1}{\sqrt{{(sec y)}^{2} - 1}} ⎞ ⎟ ⎟ ⎠$ $13/2Ln((secy)^2-1)+3Ln((secy+1)/sqrt((secy)^2-1))$

= $\frac{13}{2} ln ({(sec y)}^{2} - 1) + \frac{3}{2} ln (\frac{{(sec y + 1)}^{2}}{{(sec y)}^{2} - 1})$ $13/2Ln((secy)^2-1)+3/2Ln((secy+1)^2/((secy)^2-1))$

= $\frac{13}{2} ln ({(sec y)}^{2} - 1) + \frac{3}{2} ln (\frac{sec y + 1}{sec y - 1})$ $13/2Ln((secy)^2-1)+3/2Ln((secy+1)/(secy-1))$

After using $2 x + 13 = 3 sec y$ $2x+13=3secy$ and $sec y = \frac{2 x + 13}{3}$ $secy=(2x+13)/3$ inverse transforms, I found

$A = \frac{13}{2} {ln (\frac{2 x + 13}{3})}^{2} - 1) + \frac{3}{2} ln (\frac{\frac{2 x + 13}{3} + 1}{\frac{2 x + 13}{3} - 1})$ $A=13/2Ln((2x+13)/3)^2-1)+3/2Ln(((2x+13)/3+1)/((2x+13)/3-1))$

= $\frac{13}{2} ln (\frac{4 x^{2} + 52 x + 160}{9}) + \frac{3}{2} ln (\frac{x + 8}{x + 5})$ $13/2Ln((4x^2+52x+160)/9)+3/2Ln((x+8)/(x+5))$

Thus,

$\int ln (x^{2} + 13 x + 40) \cdot d x$ $int Ln(x^2+13x+40)*dx$

= $x ln (x^{2} + 13 x + 40) - 2 x + C 1 + \frac{13}{2} ln (\frac{4 x^{2} + 52 x + 160}{9}) + \frac{3}{2} ln (\frac{x + 8}{x + 5})$ $xLn(x^2+13x+40)-2x+C1+13/2Ln((4x^2+52x+160)/9)+3/2Ln((x+8)/(x+5))$

= $x ln (x^{2} + 13 x + 40) - 2 x + \frac{13}{2} ln ((x^{2} + 13 x + 40)$ $xLn(x^2+13x+40)-2x+13/2Ln((x^2+13x+40)$ + $\frac{3}{2} ln (\frac{x + 8}{x + 5}) + C$ $3/2Ln((x+8)/(x+5))+C$
= $\frac{2 x + 13}{2} \cdot ln (x^{2} + 13 x + 40) - 2 x + \frac{3}{2} ln (\frac{x + 8}{x + 5}) + C$ $(2x+13)/2*Ln(x^2+13x+40)-2x+3/2Ln((x+8)/(x+5))+C$

Note: $C = C 1 - \frac{13}{2} \cdot ln (\frac{4}{9}) = C 1 + 13 ln (\frac{3}{2})$ $C=C1-13/2*Ln(4/9)=C1+13Ln(3/2)$

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