For the first one, let After turning the resultant product into a sum the rest follows.
For the second one, let The remaining integral can be killed with a trig. or hyperbolic substitution.
Hello, FalconPUNCH!!
Krizalid is absolutely correct . . .
Let 1 + \ln x)\,dx" alt="u \:=\:x\ln x\quad\Rightarrow\quad du \:=\:\left(x\!\cdot\!\frac{1}{x} + \ln x\right)\,dx\:=\1 + \ln x)\,dx" />
We have: .
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Now let