A few integration by parts problems

In summary: 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
kudoushinichi88
129
2

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
 

Related to A few integration by parts problems

1. What is integration by parts?

Integration by parts is a mathematical technique used to evaluate integrals that are products of two functions. It is based on the product rule of differentiation and involves breaking down an integral into two parts and applying a specific formula.

2. When is integration by parts used?

Integration by parts is typically used when the integral involves a product of two functions that cannot be easily integrated using other techniques, such as substitution or trigonometric identities.

3. How do you choose which function to differentiate and which to integrate in integration by parts?

When using integration by parts, the choice of which function to differentiate and which to integrate is based on the acronym "LIATE", which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. The function that falls first in this order is chosen to be differentiated.

4. Can integration by parts be used for definite integrals?

Yes, integration by parts can be used for both indefinite and definite integrals. In the case of definite integrals, the limits of integration are applied to the final solution after integration by parts has been performed.

5. Are there any limitations to using integration by parts?

Integration by parts may not always lead to a solution, and in some cases, it may lead to a more complicated integral. It is also important to be careful when choosing the parts to integrate and differentiate, as the wrong choice may result in an unsolvable integral.

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