Question

Integration by parts

Answer 1

\[\int\limits x ^{-2} \ln(x) dx\]

Answer 2

i this correct so far i got

u= x^-2 dv= ln(x)

du= -2x dx v= 1/x

Answer 3

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

Answer 4

Integral of Ln(x) is not 1/x and I do not know if it actually can be performed.

Answer 5

u right it isn't that was my mistake

Answer 6

\[\int\limits_{}^{} \ln (u) du = u \ln (u) - u + c\]

Answer 7

@Johnbc the integral \(\int \ln x~dx\) can be performed by parts.

@givanna618 for this problem you want \(u=\ln x\) and \(dv=x^{-2}~dx\).

Answer 8

It can be performed but it seems like it would be easier to do u = Ln (x)

Answer 9

okay so du= 1/x and v = x^-1

Answer 10

@Johnbc if using substitute like that, you get stuck and still have to do more.

Answer 11

-1/x for v i mean

Answer 12

Yes

Answer 13

so \[\int\limits x ^{-2} lnx dx = (\ln(x))(-1/x) - \int\limits(1/x)(-1/x) dx\]

@SithsAndGiggles

Answer 14

Right. Simplify that if you please

Answer 15

can someone please help think I'm stuck

Answer 16

can u show me how its done

Answer 17

You performed it correctly.

Answer 18

Now all you need to do is simplify the expression in the integral and integrate.