# Difference between revisions of "Math 22 Integration by Parts and Present Value"

## Integration by Parts

 Let ${\displaystyle u}$ and ${\displaystyle v}$ be differentiable functions of ${\displaystyle x}$.

${\displaystyle \int udv=uv-\int vdu}$


Exercises Use integration by parts to evaluation:

1) ${\displaystyle \int \ln xdx}$

Solution:
Let ${\displaystyle u=\ln x}$, ${\displaystyle >du={\frac {1}{x}}dx}$
and ${\displaystyle dv=dx}$ and ${\displaystyle v=x}$
Then, by integration by parts: ${\displaystyle \int \ln xdx=x\ln x-\int x{\frac {1}{x}}dx=x\ln x-\int dx=x\ln x-x+C}$

2) ${\displaystyle \int xe^{3x}dx}$

Solution:
Let ${\displaystyle u=x}$, ${\displaystyle du=dx}$
and ${\displaystyle dv=e^{3x}dx}$ and ${\displaystyle v={\frac {1}{3}}e^{3x}}$
Then, by integration by parts: ${\displaystyle \int xe^{3x}dx=x{\frac {1}{3}}e^{3x}-\int {\frac {1}{3}}e^{3x}dx=x{\frac {1}{3}}e^{3x}-{\frac {1}{9}}e^{3x}+C}$

3) ${\displaystyle \int x^{2}e^{-x}dx}$

Solution:
Let ${\displaystyle u=x^{2}}$, ${\displaystyle du=2xdx}$
and ${\displaystyle dv=e^{-x}dx}$ and ${\displaystyle v=-e^{-x}}$
Then, by integration by parts: ${\displaystyle \int x^{2}e^{-x}dx=x^{2}(-e^{-x})-\int -e^{-x}2xdx=-x^{2}e^{-x}+\int 2xe^{-x}dx}$
Now, we apply integration by parts the second time for ${\displaystyle \int 2xe^{-x}dx}$
Let ${\displaystyle u=2x}$, ${\displaystyle du=2dx}$
and ${\displaystyle dv=e^{-x}dx}$ and ${\displaystyle v=-e^{-x}}$
So ${\displaystyle \int 2xe^{-x}dx=2x(-e^{-x})-\int -e^{-x}2dx=-2xe^{-x}-e^{-x}+C}$
Therefore, ${\displaystyle \int x^{2}e^{-x}dx=-x^{2}e^{-x}-2xe^{-x}-e^{-x}+C}$

## Note

1. Tabular method is convenient in some cases.