Integration Q (1 Viewer)

[Kipling]

How do you integrate ln(x) / x^4 from 1 to e? Thanks in advance

 

[HeroicPandas]

As asianese said in the 4u marathon, IBP (integration by parts)

let u = ln(x) and dv = 1/x^4 dx

blah blah

 

[Kipling]

Another question: Using the f(x) = f(a-x) rule regarding integration, prove that (xsin(2x))/(1+cos(2x)) from 0 to pi/2 equals (pi^2)/16

Also: <a href="http://www.codecogs.com/eqnedit.php?latex=\mathrm{If} \; \; I_{n}= \int sec^nx, \mathrm\; \mathrm{{show\: that }}\; I_{n}= \frac{1}{n-1}(tanxsec^{n-2})@plus;\frac{n-2}{n-1}I_{n-2} \; \; \; \; \; \mathrm{Hence \: evaluate\: }I_{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\mathrm{If} \; \; I_{n}= \int sec^nx, \mathrm\; \mathrm{{show\: that }}\; I_{n}= \frac{1}{n-1}(tanxsec^{n-2})+\frac{n-2}{n-1}I_{n-2} \; \; \; \; \; \mathrm{Hence \: evaluate\: }I_{4}" title="\mathrm{If} \; \; I_{n}= \int sec^nx, \mathrm\; \mathrm{{show\: that }}\; I_{n}= \frac{1}{n-1}(tanxsec^{n-2})+\frac{n-2}{n-1}I_{n-2} \; \; \; \; \; \mathrm{Hence \: evaluate\: }I_{4}" /></a>

 

[Makematics]

Another question: Using the f(x) = f(a-x) rule regarding integration, prove that (xsin(2x))/(1+cos(2x)) from 0 to pi/2 equals (pi^2)/16

Also: <a href="http://www.codecogs.com/eqnedit.php?latex=\mathrm{If} \; \; I_{n}= \int sec^nx, \mathrm\; \mathrm{{show\: that }}\; I_{n}= \frac{1}{n-1}(tanxsec^{n-2})@plus;\frac{n-2}{n-1}I_{n-2} \; \; \; \; \; \mathrm{Hence \: evaluate\: }I_{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\mathrm{If} \; \; I_{n}= \int sec^nx, \mathrm\; \mathrm{{show\: that }}\; I_{n}= \frac{1}{n-1}(tanxsec^{n-2})+\frac{n-2}{n-1}I_{n-2} \; \; \; \; \; \mathrm{Hence \: evaluate\: }I_{4}" title="\mathrm{If} \; \; I_{n}= \int sec^nx, \mathrm\; \mathrm{{show\: that }}\; I_{n}= \frac{1}{n-1}(tanxsec^{n-2})+\frac{n-2}{n-1}I_{n-2} \; \; \; \; \; \mathrm{Hence \: evaluate\: }I_{4}" /></a>

For the second part, it should be a simple IBP. split the integral into sec^(n-2)x and sec^2x and then let u=sec^(n-2)x, v=tanx, and proceed.