E-Store
[1.6]Indefinite Integral
|
| Forum Index -> Integral Calculus -> View Full Question |
|
| Author | Message | |||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
xlog(x+1) dx = ln(x+1)  x dx - x2/2 . 1/(x+1) dx [ integrating by parts ]= x2 / [2.ln(x+1) ] - 1/2 [ (x2+x - (x+1) + 1] / (x+1) dx= x2 / [2.ln(x+1) ] - 1/2 x dx + 1/2 dx - 1/(x+1) dx=x2 / [2.ln(x+1) ] - x2 / 4 + x/2 - ln (x+1) + C Cheers !! |
|||||||||||||
You never know what is enough till you know what is more than enough. Titun |
||||||||||||||
| Like 0 people liked this | ||||||||||||||
|
|
||||||||||||||
Free Sign Up!
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
2358 Students Currently Online |
Connect with Us  |
 |
Vriti Communities
Privacy Policy © Vriti 2011