Integral Problem

[grapz]

If k is a constant such that the [ integral from ln 2 --> 0 of xe^(2x) / ( 1 + 2x)^2 ] = k / ( 4 + 8ln2)

then k is equal to

a : 3 - 2 ln 2
b: 4 - ln 2
c: 4- 3ln2

answer is a.

I tried to do integration by parts, but i don't seem to be getting anywhere.

 

[o_O]

\(\displaystyle \int_{\ln2}^{0} \frac{xe^{2x}}{(1+2x)^{2}}dx = \frac{k}{4 + 8\ln2}\)

Looking at the integral, let u = 2x + 1:

\(\displaystyle \begin{array}{ccc}u = 2x + 1 & \Rightarrow & du = 2dx \\ x = \frac{1}{2}(u-1) & & dx = \frac{1}{2}du\)

Resub:

\(\displaystyle \int_{\ln 2}^{0}\frac{xe^{2x}}{(1+2x)^{2}}dx = \int_{u\left(\ln2\right)}^{u(0)} \frac{\frac{1}{2}(u-1)e^{u-1}}{u^{2}} \cdot \frac{1}{2}du\)

\(\displaystyle = \frac{1}{4} \int_{u\left(\ln2\right)}^{u(0)} \frac{(u-1)e^{u-1}}{u^{2}}du\)

\(\displaystyle = - \frac{1}{4} \int_{u(0)}^{u\left(\ln2\right)} \frac{(u-1)e^{u-1}}{u^{2}}du\)

\(\displaystyle = -\frac{1}{4} \int_{u(0)}^{u\left(\ln2\right)} \frac{ue^{u-1} - e^{u-1}}{u^{2}}du\)

Doesn't this look familiar to a quotient rule for derivatives ?

Should be simple algebra from here to find k.

 

[markbrock30]

What do you use to draw your math equation pictures?

 

[galactus]

That's LaTex and you don't have to use anything.Just type in the code. Click on 'quote' to see what was used to make it display that way.