Question
Evaluate the integral
ln(2)
Alternative Form
≈0.693147
Evaluate
∫e2e3x(ln(x)−1)1dx
Multiply the terms
More Steps
Evaluate
x(ln(x)−1)
Use the the distributive property to expand the expression
xln(x)+x(−1)
Multiplying or dividing an odd number of negative terms equals a negative
xln(x)−x
∫e2e3xln(x)−x1dx
Evaluate the integral
∫xln(x)−x1dx
Rewrite the expression
∫x(ln(x)−1)1dx
Use the substitution dx=xdt to transform the integral
More Steps
Evaluate
t=ln(x)
Calculate the derivative
dt=x1dx
Evaluate
dx=xdt
∫x(ln(x)−1)1×xdt
Cancel out the common factor x
∫ln(x)−11dt
Use the substitution t=ln(x) to transform the integral
∫t−11dt
Use the property of integral ∫ax+b1dx=a1ln∣ax+b∣
ln(∣t−1∣)
Substitute back
ln(∣ln(x)−1∣)
Return the limits
(ln(∣ln(x)−1∣))e2e3
Solution
More Steps
Substitute the values into formula
ln(ln(e3)−1)−ln(ln(e2)−1)
Use lnen=n to simplify the expression
ln(∣3−1∣)−ln(ln(e2)−1)
Use lnen=n to simplify the expression
ln(∣3−1∣)−ln(∣2−1∣)
Subtract the numbers
ln(∣2∣)−ln(∣2−1∣)
Subtract the numbers
ln(∣2∣)−ln(∣1∣)
When the expression in absolute value bars is not negative, remove the bars
ln(2)−ln(∣1∣)
When the expression in absolute value bars is not negative, remove the bars
ln(2)−ln(1)
Evaluate the logarithm
ln(2)−0
Removing 0 doesn't change the value,so remove it from the expression
ln(2)
ln(2)
Alternative Form
≈0.693147
Show Solution