How do i find the antiderivative of this

[afcwestwarrior]

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

 

[tim_lou]

hmmm what did you do?

consider the u-substitution of
u=x^2

 

[afcwestwarrior]

so i have to use the chain rule or the quotient rule when i find the antiderivative of this

 

[afcwestwarrior]

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

 

[eigenglue]

Let u = x^2 + 1
Then du = 2x*dx
So x*dx/(x^2 + 1)^(1/2) = du/(2*u^(1/2))

 

[VietDao29]

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.
\int \frac{f(x)}{g(x)} dx \neq \frac{\int f(x) dx}{\int g(x) dx}, they are not the same.

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

Example 1:
\int x \sqrt{x ^ 2 + 3} dx

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.

\int x \sqrt{x ^ 2 + 3} dx = \int \sqrt{x ^ 2 + 3} x dx
Let u = x ^ 2 + 3 \Rightarrow du = (x ^ 2 + 3)'_x dx = 2x dx \Rightarrow x dx = \frac{du}{2}
The integral becomes:
... = \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

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

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

Example 2:
\int x e ^ {3x ^ 2 + 5} dx

It's pretty similar to the previous one, let u = 3x ^ 2 + 5 \Rightarrow du = 6x dx \Rightarrow x dx = \frac{du}{6}
The integral becomes:
\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

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