Question

integrate 1/(x^2+x)

Answer 1

\[\int\frac{1}{x^2+x}dx\]
Complete the square in the denominator:
\[x^2+x\\
x^2+x+\frac{1}{4}-\frac{1}{4}\\
\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\]
\[\int\frac{1}{\left(x+\frac{1}{2}\right)^2-\frac{1}{4}}dx\]
\[\text{Let }x+\frac{1}{2}=\frac{1}{2}\sec t\\
dx=\frac{1}{2}\sec t\tan t\;dt\]
\[\int\frac{\frac{1}{2}\sec t\tan t}{\left(\frac{1}{2}\sec t\right)^2-\frac{1}{4}}dt\\
\int\frac{\frac{1}{2}\sec t\tan t}{\frac{1}{4}\sec^2t-\frac{1}{4}}dt\\
2\int\frac{\sec t\tan t}{\tan^2t}dt\\
2\int\frac{\sec t}{\tan t}dt\\
2\int\frac{\frac{1}{\cos t}}{\frac{\sin t}{\cos t}}dt\\
2\int\frac{1}{\sin t}dt\\
2\int\csc t\; dt\]
and so on.