Title: Integration #2
Course: Demonstrating STACK (v4.4.7)

Evaluate the following integral, using a substitution if necessary: \[ \int \frac{1}[value]\ln\left(\frac[value][value]\right)d[value]\]
Feedback: Let \[u = \ln\left(\frac[value][value]\right)\] \[\frac{du}{d[value]}=-\frac[value][value]\frac[value][value]=-\frac{1}[value]\] Therefore \[\int\frac{1}[value]\ln\left(\frac[value][value]\right)d[value] = -\int udu = -\frac{u^2}{2} +c = -\frac{1}{2}\ln^2\left(\frac[value][value]\right) +c\]

Evaluate the following integral: \[ \int [value]d[value] \]
Feedback: Dividing [value] by [value] we get \[[value]\] Therefore \[\int[value]d[value] =\int\left([value]\right)d[value]=[value]\]

Evaluate the following integral: \[ \int [value]d[value] \]
Feedback: Let \([value]+[value]=R\), therefore \(d[value]=dR\) \[\int[value]d[value] =\int\frac[value]{R^2}dR\] \[=\int\frac[value]{R^2}dR=\int([value])dR\] \[=[value]+c=[value]\]

Evaluate the following integral, using a substitution if necessary: \[ \int [value] d[value] \]
Feedback: Let \[\frac[value][value][value]=\sin(u)\] \[\frac[value][value]d[value]=\cos(u)du\] \[du= [value]\frac{d[value]}{\cos(u)} = [value]\frac{d[value]}[value]= [value]\frac{d[value]}[value] = \frac{[value]d[value]}[value]\] Therefore \[\int [value]d[value] = \int\frac{du}[value]= \frac{u}[value] + c = \frac{1}[value][value]+c\]