Leave that red part out.chapsticks said:Homework Statement
Find the integral. (Note: Solve by the simplest method - not all require integration by parts.)
∫t ln(t + 2) dt
Homework Equations
∫udv=uv-∫vdu
The Attempt at a Solution
dv=tdt v=tdt=t2/2
u=ln(t+2) du=1/(t+2)dt
Check the vdu in that last integral.∫t ln(t + 2) dt
=(t2/2)ln(t+2)-(1/2)∫(t2/2)dt
chapsticks said:Homework Statement
Find the integral. (Note: Solve by the simplest method - not all require integration by parts.)
∫t ln(t + 2) dt
Homework Equations
∫udv=uv-∫vdu
The Attempt at a Solution
dv=tdt v=tdt=t2/2
u=ln(t+2) du=1/(t+2)dt
∫t ln(t + 2) dt
=(t2/2)ln(t+2)-(1/2)∫(t2/2)dt
=(t2/2)ln(t+2)-(1/2)∫(t-2+(4/(t+2))dt
=(t2/2)ln(t+2)-(1/2)[(t2/2)-2t+4ln(t+2)]+C
=? Idk why I keep getting it wrong?
An integral is a mathematical concept used to find the area under a curve in a graph. It is also known as anti-derivative as it is the reverse process of differentiation.
The simplest method to find an integral is by using basic integration rules such as the power rule, product rule, quotient rule, and chain rule.
A definite integral has specific limits of integration, while an indefinite integral does not. This means that a definite integral will give a numerical value, while an indefinite integral will give a function.
No, not all functions can be integrated. Some functions are not continuous or have complicated forms that do not have a simple integral form.
Finding an integral has many applications in mathematics and science, such as calculating areas, volumes, and finding the average value of a function. It is also used in physics to calculate work and in economics to calculate total revenue and profit.