How do i find the antiderivative of this

  • #1
x/ square root sign(x^2 + 1)

i did this x/ (x^2 + 1)^(1/2) simplified it

then i did this (x^2/2)/ 3/2(x^2 + 1)^(3/2) antidirevitionated it

then i cleaned it up x/ 3( x^2 + 1)^ 3

is this right
 
Last edited:

Answers and Replies

  • #2
682
1
hmmm what did you do?

consider the u-substitution of
[tex]u=x^2[/tex]
 
  • #3
so i have to use the chain rule or the quotient rule when i find the antiderivative of this
 
  • #4
verify by differentiation that the formula x/square root (x^2 + 1) + C
is correct

this is what i have to do, but first i have to find the antiderivate of that function or polynomial, and then i have to just differentiate it which is easy
 
  • #5
25
0
Let u = x^2 + 1
Then du = 2x*dx
So x*dx/(x^2 + 1)^(1/2) = du/(2*u^(1/2))
 
  • #6
VietDao29
Homework Helper
1,423
3
so i have to use the chain rule or the quotient rule when i find the antiderivative of this
No, you should remember that there's no such thing as Chain rule, or Quotient Rule when finding an anti-derivative to one function.
You should re-read your textbook again carefully, and slowly to fully understand what you have to do. In integration, we can make a u-substitution, a trig-substitution, Integration by Parts, blah, blah, blah,... but not chain rule, or quotient rule. They are just for differentiating.
What you have done in your first post is not correct.
[tex]\int \frac{f(x)}{g(x)} dx \neq \frac{\int f(x) dx}{\int g(x) dx}[/tex], they are not the same.

-----------------------

I'll give you an example, similar to the problem. But you really should read your text book first.

Example 1:
[tex]\int x \sqrt{x ^ 2 + 3} dx[/tex]

You should notice the x in front of the square root, it differs from the derivative of (x2 + 3, i.e, the expression inside the square root) by a factor 2.

[tex]\int x \sqrt{x ^ 2 + 3} dx = \int \sqrt{x ^ 2 + 3} x dx[/tex]
Let [tex]u = x ^ 2 + 3 \Rightarrow du = (x ^ 2 + 3)'_x dx = 2x dx \Rightarrow x dx = \frac{du}{2}[/tex]
The integral becomes:
[tex]... = \int \sqrt{u} \frac{du}{2} = \frac{1}{2} \int \sqrt{u} du = \frac{1}{2} \int u ^ {\frac{1}{2}} du = \frac{1}{2} \times \frac{u ^ {\frac{3}{2}}}{\frac{3}{2}} + C[/tex]

[tex]= \frac{1}{3} u ^ {\frac{3}{2}} + C = \frac{1}{3} \sqrt{u ^ 3} + C[/tex], now, change u back to x, we have:

[tex]... = \frac{1}{3} \sqrt{(x ^ 2 + 3) ^ 3} + C[/tex]

Example 2:
[tex]\int x e ^ {3x ^ 2 + 5} dx[/tex]

It's pretty similar to the previous one, let [tex]u = 3x ^ 2 + 5 \Rightarrow du = 6x dx \Rightarrow x dx = \frac{du}{6}[/tex]
The integral becomes:
[tex]\int e ^ u \frac{du}{6} = \frac{1}{6} \int e ^ u du = \frac{1}{6} e ^ u + C = \frac{1}{6} e ^ {3x ^ 2 + 5} + C[/tex]

Now, you can just do the same to your problem. Can you go from here? :)
 

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