A few integration by parts problems

For the second integral you can try:u=\ln(9+x^2) to getdu=\frac{2x}{9+x^2}dx and dv=x dx to getx\ln(9+x^2)-\int \frac{2x^2}{9+x^2}dxFor the third integral you can try:u=x^2 to getdu=2xdx and dv=\tan^{-1}x to getx^2
  • #1

Homework Statement



Hello. I am doing some problems on integration by parts and got stuck on the following problems. Any help would be appreciated.

i. [tex]\int \arcsin x dx[/tex]

ii. [tex]\int_{0}^{1} x \ln (9+x^2) dx[/tex]

iii. [tex]\int x^2 \arctan x\, dx[/tex]

Homework Equations



[tex]u\,du=uv-v\,du[/tex]

The Attempt at a Solution



i. I tried u= arcsin x and dv=dx, but ended up with an integral which looks worse than the original question.

ii. I used [itex]u= \ln (9+x^2)\,dx[/itex] but ended up with

[tex]\int_{0}^{1} \frac{x^3}{9+x^2}\,dx[/tex]
which I am stuck with.

iii. I tried u=arctan x and dv= x^2 dx but ended up with a similar looking integral in question ii, which is [tex]\int x \ln(1+x^2)\,dx[/tex]
 
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  • #2
You are getting stuck at a point where a simple substitution would finish it. Look at ii). Try the u-substitution u=9+x^2.
 
  • #3
#1: you gave up too soon. Show us what you got.

#2: don't integrate by parts yet. Perform a u-substitution first.

#3: your choice for u and dv are fine, but I don't get that integral that's similar to #2. Again, show us what you got.
 
  • #4
For 1) look at the integral:
[tex]
\int 1\cdot\sin^{-1}xdx
[/tex]
Use u=arcsin x and dv=1
 

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