# How do you integrate ln(x^2 + 13x + 40) dx?

Mar 15, 2018

$\frac{2 x + 13}{2} \cdot L n \left({x}^{2} + 13 x + 40\right) - 2 x + \frac{3}{2} L n \left(\frac{x + 8}{x + 5}\right) + C$

#### Explanation:

$\int L n \left({x}^{2} + 13 x + 40\right) \cdot \mathrm{dx}$

=$x L n \left({x}^{2} + 13 x + 40\right) - \int x \cdot \frac{\left(2 x + 13\right) \cdot \mathrm{dx}}{{x}^{2} + 13 x + 40}$

=$x L n \left({x}^{2} + 13 x + 40\right) - \int \frac{\left(2 {x}^{2} + 13 x\right) \cdot \mathrm{dx}}{{x}^{2} + 13 x + 40}$

=$x L n \left({x}^{2} + 13 x + 40\right) - \int \frac{\left(2 {x}^{2} + 26 x + 80 - 13 x - 80\right) \cdot \mathrm{dx}}{{x}^{2} + 13 x + 40}$

=$x L n \left({x}^{2} + 13 x + 40\right) - \int 2 \mathrm{dx} + \int \frac{\left(13 x + 80\right) \cdot \mathrm{dx}}{{x}^{2} + 13 x + 40}$

=$x L n \left({x}^{2} + 13 x + 40\right) - 2 x + C 1 + \int \frac{\left(52 x + 320\right) \cdot \mathrm{dx}}{4 {x}^{2} + 52 x + 160}$

=$x L n \left({x}^{2} + 13 x + 40\right) - 2 x + C 1 + \int \frac{\left(26 x + 160\right) \cdot 2 \mathrm{dx}}{{\left(2 x + 13\right)}^{2} - {3}^{2}}$

$A = \int \frac{\left(26 x + 160\right) \cdot 2 \mathrm{dx}}{{\left(2 x + 13\right)}^{2} - {3}^{2}}$

After using $2 x + 13 = 3 \sec y$, $2 \mathrm{dx} = 3 \sec y \cdot \tan y \cdot \mathrm{dy}$ and $x = \frac{3 \sec y - 13}{2}$ transforms, $A$ became

$A = \int \frac{\left(26 \cdot \frac{3 \sec y - 13}{2} + 160\right) \cdot 3 \sec y \cdot \tan y \cdot \mathrm{dy}}{9 {\left(\tan y\right)}^{2}}$

=$\int \frac{\left(39 \sec y - 9\right) \cdot \sec y \cdot \mathrm{dy}}{3 \tan y}$

=$13 \int \frac{{\left(\sec y\right)}^{2} \cdot \mathrm{dy}}{\tan y} - 3 \int \frac{\sec y \cdot \mathrm{dy}}{\tan} y$

=$13 L n \left(\tan y\right) - 3 \int \csc y \cdot \mathrm{dy}$

=$\frac{13}{2} L n \left({\left(\tan y\right)}^{2}\right) - 3 \int \frac{\csc y \cdot \left(\csc y + \cot y\right) \cdot \mathrm{dy}}{\csc y + \cot y}$

=$\frac{13}{2} L n \left({\left(\sec y\right)}^{2} - 1\right) + 3 L n \left(\csc y + \cot y\right)$

=$\frac{13}{2} L n \left({\left(\sec y\right)}^{2} - 1\right) + 3 L n \left(\frac{\sec y + 1}{\tan} y\right)$

=$\frac{13}{2} L n \left({\left(\sec y\right)}^{2} - 1\right) + 3 L n \left(\frac{\sec y + 1}{\sqrt{{\left(\sec y\right)}^{2} - 1}}\right)$

=$\frac{13}{2} L n \left({\left(\sec y\right)}^{2} - 1\right) + \frac{3}{2} L n \left({\left(\sec y + 1\right)}^{2} / \left({\left(\sec y\right)}^{2} - 1\right)\right)$

=$\frac{13}{2} L n \left({\left(\sec y\right)}^{2} - 1\right) + \frac{3}{2} L n \left(\frac{\sec y + 1}{\sec y - 1}\right)$

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

A=13/2Ln((2x+13)/3)^2-1)+3/2Ln(((2x+13)/3+1)/((2x+13)/3-1))

=$\frac{13}{2} L n \left(\frac{4 {x}^{2} + 52 x + 160}{9}\right) + \frac{3}{2} L n \left(\frac{x + 8}{x + 5}\right)$

Thus,

$\int L n \left({x}^{2} + 13 x + 40\right) \cdot \mathrm{dx}$

=$x L n \left({x}^{2} + 13 x + 40\right) - 2 x + C 1 + \frac{13}{2} L n \left(\frac{4 {x}^{2} + 52 x + 160}{9}\right) + \frac{3}{2} L n \left(\frac{x + 8}{x + 5}\right)$

=xLn(x^2+13x+40)-2x+13/2Ln((x^2+13x+40)+$\frac{3}{2} L n \left(\frac{x + 8}{x + 5}\right) + C$
=$\frac{2 x + 13}{2} \cdot L n \left({x}^{2} + 13 x + 40\right) - 2 x + \frac{3}{2} L n \left(\frac{x + 8}{x + 5}\right) + C$

Note: $C = C 1 - \frac{13}{2} \cdot L n \left(\frac{4}{9}\right) = C 1 + 13 L n \left(\frac{3}{2}\right)$