Question

Please help me get started with solving the following integral (click to see).

Answer 1

\[\int\limits_{}^{}(lnx)^{2}dx\]

Answer 2

\[\int\limits_{}^{}u ^{2}\frac{ du }{ 1/x }\]
Is this the correct start before doing integration by parts?

Answer 3

u=ln[x]^2, v=x, du=2ln[x]1/x Integration by parts: uv-integral[vdu]

Answer 4

uv-integral[vdu]= xln[x]^2-integral[2xln[x]1/x]= xln[x]^2-2integral[ln[x]]

Answer 5

questions or you got it from here? You will do another integration by parts. So a total of two integration by parts to get to a simple 1dx integral.

Answer 6

step 1: \[xln[x]^{2}-\int\limits_{}^{}2x \ln [x] \frac{ 1 }{ x }\]

Answer 7

simplify:
\[xln[x]^2-2\int\limits_{}^{}\ln[x]dx \]

Answer 8

\[xln[x]^2-2(xln[x]-\int\limits_{}^{}x(1/x)dx=xln[x]^2-2(xln[x]-\int\limits_{}^{}1dx\]

Answer 9

\[x(\ln[x]^2-2\ln[x]+2)+c\]

Answer 10

What do you think? I can show you an easy way to remember these multi step problems if you'd like....