# Thread: Integration by Parts

1. ## Integration by Parts

Hello guys, I've got a math quiz tomorrow. My professor only uses previous homework problems that he assigns so I'm just going over all of the problems in the book that he assigned, lol. I'm having trouble with this question:

For the following problem, use integration by parts with the suggested u and dv to find the antiderivative.

$\displaystyle {\int_{}\frac{lnx}{x^2}} dx$

So... find the antiderivitive of: (lnx/x^2) dx ... u = lnx dv = (x^-2)dx

I get this far:
u = lnx
du/dx = 1/x
du = 1/x dx

dv = x^-2dx
dv/dx = x^-2
v = x^-1/1

So... u * v - integral (v * du)

lnx * x^-1/1 - integral (x^-1/1 * 1/x) dx

2. Originally Posted by larson
For the following problem, use integration by parts with the suggested u and dv to find the antiderivative.

$\displaystyle {\int_{}\frac{lnx}{x^2}} dx$
Let $\displaystyle f(x)=\frac{\ln x}x\implies f'(x)=\frac{1-\ln x}{x^2}.$

After integration we have $\displaystyle \frac{{\ln x}} {x} + k_1 = - \frac{1} {x} - \int {\frac{{\ln x}} {{x^2 }}\,dx} \,\therefore \,\int {\frac{{\ln x}} {{x^2 }}\,dx} = - \frac{1} {x} - \frac{{\ln x}} {x} + k.$

3. Originally Posted by Krizalid
Let $\displaystyle f(x)=\frac{\ln x}x\implies f'(x)=\frac{1-\ln x}{x^2}.$

After integration we have $\displaystyle \frac{{\ln x}} {x} + k_1 = - \frac{1} {x} - \int {\frac{{\ln x}} {{x^2 }}\,dx} \,\therefore \,\int {\frac{{\ln x}} {{x^2 }}\,dx} = - \frac{1} {x} - \frac{{\ln x}} {x} + k.$
hey, I'm still a little confused... you didn't use integration by parts did you?

4. Originally Posted by larson
Hello guys, I've got a math quiz tomorrow. My professor only uses previous homework problems that he assigns so I'm just going over all of the problems in the book that he assigned, lol. I'm having trouble with this question:

For the following problem, use integration by parts with the suggested u and dv to find the antiderivative.

$\displaystyle {\int_{}\frac{lnx}{x^2}} dx$

So... find the antiderivitive of: (lnx/x^2) dx ... u = lnx dv = (x^-2)dx

I get this far:
u = lnx
du/dx = 1/x
du = 1/x dx

dv = x^-2dx
dv/dx = x^-2
v = x^-1/1

So... u * v - integral (v * du)

lnx * x^-1/1 - integral (x^-1/1 * 1/x) dx
Krizalid's way is nice and easy, as usual (oy!) -- no, Krizalid did not use integration by parts. he literally reversed the derivative by integrating both sides of the derivative equation.

but if you insist on doing integration by parts, i'd do a substitution first

Let $\displaystyle t = \ln x \implies e^t = x$

$\displaystyle \Rightarrow dt = \frac 1x~dx$

So our integral becomes:

$\displaystyle \int te^{-t}~dt$

which is a lot less intimidating to do integration by parts on

of course, your original approach may work, i didn't check it. this is just the way i saw

5. Or integrate by parts as it's written:

By setting $\displaystyle u=\ln x\implies du=\frac1x\,dx$ & $\displaystyle dv=x^{-2}\,dx\implies v=-\frac1x,$ your integral becomes

$\displaystyle \int {\frac{{\ln x}} {{x^2 }}\,dx} = - \frac{{\ln x}} {x} + \int {\frac{1} {{x^2 }}\,dx},$ the rest follows.

6. Originally Posted by Krizalid
Or integrate by parts as it's written:

By setting $\displaystyle u=\ln x\implies du=\frac1x\,dx$ & $\displaystyle dv=x^{-2}\,dx\implies v=-\frac1x,$ your integral becomes

$\displaystyle \int {\frac{{\ln x}} {{x^2 }}\,dx} = - \frac{{\ln x}} {x} + \int {\frac{1} {{x^2 }}\,dx},$ the rest follows.
Now I understand, thank you both Krizalid and Jhevon!