U substitution

**mike21** · June 8th 2010, 10:32 AM

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.

**skeeter** · June 8th 2010, 10:41 AM

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.

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

$u = \ln(2x+3)<br />$

$du = \frac{2}{2x+3} \, dx<br />$

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

substitute and reset limits ...

$\frac{1}{2} \int_{\ln{4}}^{\ln{5}} u \, du<br />$

finish it.

**redsoxfan325** · June 8th 2010, 10:42 AM

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 $\frac{1}{2}\int_{\ln4}^{\ln5}u\,du$

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