Given:
[tex]\int_4^{\infty}\frac{1}{x^2+x}\text{ dx}[/tex]To find:
the integral
[tex]\begin{gathered} First,\text{ we will re-write the expression} \\ \frac{1}{x^2+x}\text{ = }\frac{1}{x^2(1\text{ + }\frac{1}{x})} \\ \\ let\text{ u = 1 + 1/x} \\ u\text{ = 1 + x}^{-1} \\ \frac{du}{dx\text{ }}\text{ = 0 + \lparen-1}x^{-1-1})\text{ = -1x}^{-2} \end{gathered}[/tex][tex]\begin{gathered} \frac{du}{dx}\text{ = -x}^{-2} \\ \\ du\text{ = -x}^{-2}dx \\ du\text{ = }\frac{dx}{-x^2} \\ \\ \int_4^{\infty}\frac{1}{x^2+x\text{ }}dx\text{ = }\int_4^{\infty}\frac{1}{x^2(1\text{ +}\frac{1}{x})}dx \\ \\ Substitute\text{ for u and du in the expression:} \\ \int_4^{\infty}\frac{1}{x^2(u)}dx\text{ = }\int_4^{\infty}\frac{dx}{-x^2(u)}=\int_4^{\infty}-\frac{du}{u} \\ \end{gathered}[/tex][tex]\begin{gathered} -\int_4^{\infty}\frac{du}{u}=-\int_4^{\infty}ln\text{ u \lparen differentiation rule\rparen} \\ \\ \int_4^{\infty}ln(1+\frac{1}{x})=\int_4^{\infty}ln(\frac{x+1}{x})=\int_4^{\infty}ln(x+1)\text{ - ln\lparen x\rparen} \\ \\ -\int_4^{\infty}ln(1+\frac{1}{x})=-\int_4^{\infty}ln(x+1)\text{ - ln\lparen x\rparen = }\int_4^{\infty}ln(x)\text{ - ln\lparen x+1\rparen} \\ \\ -\int_4^{\infty}ln(1+\frac{1}{x})=\text{ \lbrack\lparen}\lim_{x\to\infty}(ln(x)\text{ - ln\lparen x+1\rparen\rbrack- \lbrack lnx - ln\lparen x+1\rparen\rbrack}_{x=4} \\ \\ -\int_4^{\infty}ln(1+\frac{1}{x})=\text{ \lbrack}\frac{x}{x+1}\text{\rbrack}_{\infty}\text{ - ln\lbrack}\frac{x}{x+1}]_4 \\ \\ -\int_4^{\infty}ln(1+\frac{1}{x})=0\text{ - ln\lbrack}\frac{4}{4+1}] \\ \\ -\int_4^{\infty}ln(1+\frac{1}{x})=\text{ -ln\lbrack}\frac{4}{5}] \end{gathered}[/tex][tex]-\int_4^{\infty}ln(1+\frac{1}{x})\text{ = ln\lparen}\frac{5}{4})[/tex]