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

## Integration by Parts

 Let $u$ and $v$ be differentiable functions of $x$ .

$\int udv=uv-\int vdu$ Exercises Use integration by parts to evaluation:

1) $\int \ln xdx$ Solution:
Let $u=\ln x$ , $>du={\frac {1}{x}}dx$ and $dv=dx$ and $v=x$ Then, by integration by parts: $\int \ln xdx=x\ln x-\int x{\frac {1}{x}}dx=x\ln x-\int dx=x\ln x-x+C$ 2) $\int xe^{3x}dx$ Solution:
Let $u=x$ , $du=dx$ and $dv=e^{3x}dx$ and $v={\frac {1}{3}}e^{3x}$ Then, by integration by parts: $\int xe^{3x}dx=x{\frac {1}{3}}e^{3x}-\int {\frac {1}{3}}e^{3x}dx=x{\frac {1}{3}}e^{3x}-{\frac {1}{9}}e^{3x}+C$ 3) $\int x^{2}e^{-x}dx$ Solution:
Let $u=x^{2}$ , $du=2xdx$ and $dv=e^{-x}dx$ and $v=-e^{-x}$ Then, by integration by parts: $\int x^{2}e^{-x}dx=x^{2}(-e^{-x})-\int -e^{-x}2xdx=-x^{2}e^{-x}+\int 2xe^{-x}dx$ Now, we apply integration by parts the second time for $\int 2xe^{-x}dx$ Let $u=2x$ , $du=2dx$ and $dv=e^{-x}dx$ and $v=-e^{-x}$ So $\int 2xe^{-x}dx=2x(-e^{-x})-\int -e^{-x}2dx=-2xe^{-x}-e^{-x}+C$ Therefore, $\int x^{2}e^{-x}dx=-x^{2}e^{-x}-2xe^{-x}-e^{-x}+C$ 4) $\int {\frac {1}{x(lnx)^{3}}}dx$ Solution:
Let $u=x$ , $du=dx$ and $dv=e^{3x}dx$ and $v={\frac {1}{3}}e^{3x}$ Then, by integration by parts: $\int xe^{3x}dx=x{\frac {1}{3}}e^{3x}-\int {\frac {1}{3}}e^{3x}dx=x{\frac {1}{3}}e^{3x}-{\frac {1}{9}}e^{3x}+C$ 