April 22nd, 2018, 12:45 AM  #1 
Senior Member Joined: Nov 2011 Posts: 250 Thanks: 3  Integral
What is the integral of the mathematical expression in the picture:

April 22nd, 2018, 12:53 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra 
That's an inequality. You can't integrate an inequality. However, we can integrate the left hand expression of the inequality by a substitution of $u=x+1$. Alternatively, \begin{align*}\frac{x^2}{x+1} &= \frac{(x+1)^2  2x  1}{x+1} \\ &= \frac{(x+1)^2  2(x+1) + 1}{x+1} \\ &= (x+1)  2 + \frac1{x+1} \\ &= x  1 +\frac1{x+1}\end{align*} Last edited by v8archie; April 22nd, 2018 at 01:02 AM. 
April 22nd, 2018, 12:58 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 21,128 Thanks: 2337  
April 22nd, 2018, 01:05 AM  #4 
Senior Member Joined: Nov 2011 Posts: 250 Thanks: 3 
This is my solution. Is it wrong? 
April 22nd, 2018, 01:13 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
The fact that f(x)< 1 does NOT mean that $\displaystyle \int f(x)dx< 1$! For one thing, a indefinite integral like this always includes an arbitrary constant. The integral can be larger or less than 1 depending upon that constant. For a definite integral it is true that if $\displaystyle f(x)< 1$ then $\displaystyle \int_a^b f(x)dx< b a$. 
April 22nd, 2018, 01:40 AM  #6  
Senior Member Joined: Nov 2011 Posts: 250 Thanks: 3 
And if the f(x) < 2, is it the same? Can you give some examples for: Quote:
 

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