# Integral Questions

• Aug 15th 2007, 07:07 AM
polymerase
Integral Questions
How do u integrate:

$
\int {x sec^{ 2 } x \,dx}
$

$
\int_{1}^{e} \dfrac{ln \frac{1}{x}}{x}
$

$
\int_{0}^{1} {x 3^{ x } \,dx}
$

$
\int \dfrac{ln \frac{1}{x}}{x^{3}}
$

Help on any would be greatly appreciated :).
• Aug 15th 2007, 07:15 AM
CaptainBlack
Quote:

Originally Posted by polymerase
How do u integrate:

$
\int {x sec^{ 2 } x \,dx}
$

$
\int_{1}^{e} \dfrac{ln \frac{1}{x}}{x}
$

$
\int_{0}^{1} {x 3^{ x } \,dx}
$

$
\int \dfrac{ln \frac{1}{x}}{x^{3}}
$

Help on any would be greatly appreciated :).

You need to use integration by parts. General rule is to put u equal to the bit
of the integrand that you can differentiate most easily (of the part that you can't integrate).

So in the first you can differentiate $x$ easily and integrate $\sec^2(x)$ easily, so you would use $u=x, dv=\sec^2(x)$

RonL
• Aug 15th 2007, 07:43 AM
polymerase
Quote:

Originally Posted by CaptainBlack
You need to use integration by parts. General rule is to put u equal to the bit
of the integrand that you can differentiate most easily (of the part that you can't integrate).

So in the first you can differentiate $x$ easily and integrate $\sec^2(x)$ easily, so you would use $u=x, dv=\sec^2(x)$

RonL

For the 2nd question i yet u = x and dv = 1/x and when i integrate i end up with:

= xlnx - \int ln x dx
= x lnx - (xlnx -x)
= x
but thats not the right answer. any recommendation?
• Aug 15th 2007, 08:08 AM
topsquark
Quote:

Originally Posted by polymerase
$
\int_{1}^{e} \dfrac{ln \frac{1}{x}}{x}
$

Remember, your $u dv$ must equal the integrand!

Let
$u = ln \left ( \frac{1}{x} \right ) \implies du = -\frac{dx}{x}$
$dv = \frac{dx}{x} \implies v = ln(x)$

So
$\int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \int \frac{ln(x)}{x} dx$

Do integration by parts on the second integral:
$\int \frac{ln(x)}{x} dx$
Let
$u = ln(x) \implies du = \frac{dx}{x}$
$dv = \frac{dx}{x} \implies v = ln(x)$

So
$\int \frac{ln(x)}{x} dx = ln^2(x) - \int \frac{ln(x)}{x}dx$

Thus
$2 \int \frac{ln(x)}{x} dx = ln^2(x)$

$\int \frac{ln(x)}{x} dx = \frac{ln^2(x)}{2}$

So back to the original integral:
$\int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \int \frac{ln(x)}{x} dx$

$\int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \frac{ln^2(x)}{2} + C$

I'll let you put the limits in.

-Dan
• Aug 15th 2007, 08:26 AM
polymerase
Quote:

Originally Posted by topsquark
Remember, your $u dv$ must equal the integrand!

Let
$u = ln \left ( \frac{1}{x} \right ) \implies du = -\frac{dx}{x}$
$dv = \frac{dx}{x} \implies v = ln(x)$

So
$\int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \int \frac{ln(x)}{x} dx$

Do integration by parts on the second integral:
$\int \frac{ln(x)}{x} dx$
Let
$u = ln(x) \implies du = \frac{dx}{x}$
$dv = \frac{dx}{x} \implies v = ln(x)$

So
$\int \frac{ln(x)}{x} dx = ln^2(x) - \int \frac{ln(x)}{x}dx$

Thus
$2 \int \frac{ln(x)}{x} dx = ln^2(x)$

$\int \frac{ln(x)}{x} dx = \frac{ln^2(x)}{2}$

So back to the original integral:
$\int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \int \frac{ln(x)}{x} dx$

$\int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \frac{ln^2(x)}{2} + C$

I'll let you put the limits in.

-Dan

Thank You
• Aug 15th 2007, 10:20 AM
galactus
Parts:

$\int{x\cdot{3^{x}}}dx$

Let $u=x, \;\ dv=3^{x}, \;\ du=dx, \;\ v=\frac{3^{x}}{ln(3)}$

$uv-\int{vdu}$

$\frac{x\cdot{3^{x}}}{ln(3)}-\frac{1}{ln(3)}\int{3^{x}}dx$

$\frac{x\cdot{3^{x}}}{ln(3)}-\frac{3^{x}}{(ln(3))^{2}}=\boxed{\frac{3^{x}}{ln(3 )}\left[x-\frac{1}{ln(3)}\right]}$
• Aug 15th 2007, 11:22 AM
Soroban
Hello, polymerase!

Here's the second one . . .

Quote:

$\int_{1}^{e} \frac{\ln(\frac{1}{x})}{x}\,dx$
The numerator is: . $\ln\left(\frac{1}{x}\right) \:=\:\ln\left(x^{-1}\right) \:=\:-\ln(x)$

The integral becomes: . $-\int\ln(x)\cdot\frac{dx}{x}$

. . Let: $u \,=\,\ln x\quad\Rightarrow\quad du \,=\,\frac{dx}{x}$

Substitute: . $-\int u\,du \:=\:-\frac{1}{2}u^2 + C$

Back-substitute: . $-\frac{1}{2}\left[\ln(x)\right]^2 + C$