# U substitution

• Jun 8th 2010, 11:32 AM
mike21
U substitution
Still confused with u substitution. Can someone show me how they would solve this by U substitution?

the integral from .5 to 1 of ln(2x+3) over 2x+3 dx.

u= ln(2x+3)
du=2/2x+3 dx

Where do i go from here? I was just puting the u on top so it would look like this... u/2x+3 but now i have a u and an x. What do I do?

Thanks.
• Jun 8th 2010, 11:41 AM
skeeter
Quote:

Originally Posted by mike21
Still confused with u substitution. Can someone show me how they would solve this by U substitution?

the integral from .5 to 1 of ln(2x+3) over 2x+3 dx.

u= ln(2x+3)
du=2/2x+3 dx

Where do i go from here? I was just puting the u on top so it would look like this... u/2x+3 but now i have a u and an x. What do I do?

Thanks.

$\int_{0.5}^1 \ln(2x+3) \cdot \frac{1}{2x+3} \, dx
$

$u = \ln(2x+3)
$

$du = \frac{2}{2x+3} \, dx
$

$\frac{1}{2} \int_{0.5}^1 \ln(2x+3) \cdot \frac{2}{2x+3} \, dx$

substitute and reset limits ...

$\frac{1}{2} \int_{\ln{4}}^{\ln{5}} u \, du
$

finish it.
• Jun 8th 2010, 11:42 AM
redsoxfan325
Quote:

Originally Posted by mike21
Still confused with u substitution. Can someone show me how they would solve this by U substitution?

the integral from .5 to 1 of ln(2x+3) over 2x+3 dx.

u= ln(2x+3)
du=2/2x+3 dx

Where do i go from here? I was just puting the u on top so it would look like this... u/2x+3 but now i have a u and an x. What do I do?

Thanks.

$\int_{1/2}^1\frac{\ln(2x+3)}{2x+3}\,dx$

You are correct that

$u= \ln(2x+3)$
$du=\frac{2}{2x+3}\,dx$

Notice that $\frac{1}{2}u\,du$ is the original expression.

So you have $\int\frac{1}{2}u\,du$

To determine the bounds, simply plug them into the equation for $u$:

Lower is $\ln(2\cdot0.5 + 3) = \ln4$
Upper is $\ln(2\cdot1+3)=\ln5$

So the integral that you need to evaluate is $\frac{1}{2}\int_{\ln4}^{\ln5}u\,du$