Integration by Evaluation

Integration by Substitution/Replacement

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1. \( \int \frac{dx}{\text{Quadratic}} \)
2. \( \int \frac{\text{linear}}{\text{Quadratic}} dx \)
3. \( \int \frac{dx} {\sqrt{\text{Quadratic}}} dx \)
4. \( \int \frac{\text{linear} }{\sqrt{\text{Quadratic}}} dx \)

1. \( \int \frac{dx}{9x^2-12x+8} \)
   Solution
   Use formula \( \int \frac{dx}{x^2+a^2} dx =\frac{1}{a} \tan^{-1} \left ( \frac{x}{a}\right )+c \)
   

   \( \int \frac{dx}{9x^2-12x+8} \)	\( =\frac{1}{9}\int \frac{dx}{x^2-\frac{4}{3}x+\frac{8}{9}} \)
   	\(=\frac{1}{9}\int \frac{dx}{x^2-\frac{4}{3}x+\frac{8}{9}} \)
   	\(=\frac{1}{9}\int \frac{dx}{ \left ( x- \frac{2}{3} \right )^2+ \left (\frac{2}{3} \right )^2} \)
   	\(=\frac{1}{9} \frac{1}{\frac{2}{3}} \tan^{-1} \left ( \frac{x- \frac{2}{3}}{\frac{2}{3}}\right )+c \)
   	\(=\frac{1}{6} \tan^{-1} \left ( \frac{3x-2}{2}\right )+c \)

2. \( \int \frac{5^x}{1+5^x-5^{2x}} \)
   Solution
   Put \(5^x=u\) then \(5^x dx=\frac{1}{\log 5}du\)
   Use formula \( \int \frac{dx}{x^2-a^2} dx =\frac{1}{2a} \log \left | \frac{x-a}{x+a}\right |+c \)
   

   \( \int \frac{5^x}{1+5^x-5^{2x}} \)	\( =\frac{1}{\log 5} \int \frac{du}{1+u-u^2}\)
   	\(= -\frac{1}{\log 5} \int \frac{du}{u^2-u-1}\)
   	\(=-\frac{1}{\log 5} \int \frac{du}{ \left (u-\frac{1}{2} \right )^2- \left (\frac{\sqrt{5}}{2} \right )^2} \)

3. \(\int \frac{3x+2}{x^2+4x-1} dx\)
   Solution
   Use formula \( \int \frac{dx}{x^2-a^2} dx =\frac{1}{2a} \log \left | \frac{x-a}{x+a}\right |+c \)
   

   \(\int \frac{3x+2}{x^2+4x-1} dx\)	\( =\frac{3}{2}\int \frac{2x+\frac{4}{3}}{x^2+4x-1} dx\)
   	\(= \frac{3}{2} \left [ \int \frac{2x+(4-\frac{8}{3})}{x^2+4x-1} dx \right ]\)
   	\(=\frac{3}{2} \left [ \int \frac{2x+4}{x^2+4x-1} dx - \frac{8}{3} \int \frac{1}{x^2+4x-1} dx \right ] \)
   	\(=\frac{3}{2} \left [ \log |x^2+4x-1|- \frac{8}{3} \int \frac{1}{ (x+2)^2 -(\sqrt{5})^2 } dx \right ]+c\)

4. \(\int \frac{5x+3}{x^2+4x+10} dx= \frac{5}{2} \left [\int \frac{2x+\frac{6}{5}}{x^2+4x+10} dx \right ] = \frac{5}{2} \left [\int \frac{2x+ 4-\frac{14}{5}}{x^2+4x+10} dx \right ] = \frac{5}{2} \left [\int \frac{2x+ 4}{x^2+4x+10} dx- \frac{14}{5} \int \frac{1}{x^2+4x+10} dx \right ] \)
5. \(\int \frac{1}{2x-x^2} dx=\int \frac{1}{1^2-(x-1)^2} dx\)