integration via parts (1 Viewer)

Deem_Skills · Nov 13, 2022

Screen Shot 2022-11-13 at 9.47.48 pm.png

how to do b? Idk how to manipulate part a into b.

idkkdi · Nov 13, 2022

Deem_Skills said:
View attachment 37003
how to do b? Idk how to manipulate part a into b.

by parts b) and then u'll see

Drongoski · Nov 13, 2022

a)

$\int xlnxdx = \frac{1}{2} \int lnxdx^2 = \frac{1}{2} x^2 lnx - \frac {1}{2} \int x^2 dlnx = \cdots -\frac{1}{2} \int x^2 \times \frac {1}{x}dx \\ \\ = \frac {1}{2}x^2lnx - \frac {1}{4} x^2 + C$

b)

$\int x(ln x)^2dx = \frac {1}{2} \int (ln x)^2dx^2 = \frac {1}{2}x^2(ln x)^2 - \frac {1}{2} \int x^2\times 2lnx \times \frac{1}{x}dx \\ \\ = \cdots - \int xlnx dx = \frac {1}{2}x^2( lnx)^2 -\frac {1}{2}x^2ln x + \frac {1}{4}x^2 + D$

Some may find it hard to follow my method. But I'm literally using the form:

$\int udv = uv - \int vdu$

chilli 412 · Nov 13, 2022

the hence find in part b i assume wants us to use part a

but i dont know exactly how i wud do that any help appreciated

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integration via parts (1 Viewer)

Deem_Skills

Member

idkkdi

Well-Known Member

Drongoski

Well-Known Member

chilli 412

oo la la

the hence find in part b i assume wants us to use part a

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