- #1

- 86

- 0

**(x^2)/((x^2+1)^2)**is

__(1/2)(arctan(x)-(x/x^2+1))__

In class, we've seen the steps to solve this integral, but I don't understand certain steps..

Can someone explain me how to solve this integral, step by step?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Alexx1
- Start date

- #1

- 86

- 0

In class, we've seen the steps to solve this integral, but I don't understand certain steps..

Can someone explain me how to solve this integral, step by step?

- #2

uart

Science Advisor

- 2,776

- 9

BTW. The easiest way to do that one is "integration by parts". Have you learnt this technique yet?

- #3

- 86

- 0

BTW. The easiest way to do that one is "integration by parts". Have you learnt this technique yet?

Sure, no problem, here are the steps:

Integral((x^2)/((x^2+1)^2)dx)

= (1/2)*Integral(x d(1/(x^2+1))

= (1/2)*(x/(x^2+1))-(1/2)*Integral(1/(x^2+1)dx)

= (1/2)*(x/(x^2+1))-(arctan(x))/2

Last step is the answer

(The answer I said earlier was wrong, this is the correct answer:

Thank you

- #4

- 867

- 0

Using ∫u v' = uv - ∫v u',

let u = x and v' = x/(x

- #5

- 4

- 0

It's basically separating it into parts ie.

[itex]\int \frac{x^2}{(x^2+1)^2}\rightarrow \int \frac{x}{1}.\frac{x}{(x^2+1)^2}\equiv x(x. \sin(\arctan(x)))[/itex]

as

[itex]\frac{x}{1}=\frac{1}{2}x^2[/itex]

and [itex]x\frac{x}{(1+x)^2}=x.\sin(\arctan(x))[/itex]

By the trig identity.

Thus the answer is:

[itex]\int\frac{x^2}{(x^2+1)^2}=-\frac{1}{2}.\frac{x}{(x^2+1)}+\frac{1}{2}\arctan(x)+C[/itex]

Don't forget the constant of integration, it's a silly way to loose marks.

[itex]\int \frac{x^2}{(x^2+1)^2}\rightarrow \int \frac{x}{1}.\frac{x}{(x^2+1)^2}\equiv x(x. \sin(\arctan(x)))[/itex]

as

[itex]\frac{x}{1}=\frac{1}{2}x^2[/itex]

and [itex]x\frac{x}{(1+x)^2}=x.\sin(\arctan(x))[/itex]

By the trig identity.

Thus the answer is:

[itex]\int\frac{x^2}{(x^2+1)^2}=-\frac{1}{2}.\frac{x}{(x^2+1)}+\frac{1}{2}\arctan(x)+C[/itex]

Don't forget the constant of integration, it's a silly way to loose marks.

Last edited:

- #6

- 86

- 0

Thank you both!

- #7

- 4

- 0

Thank you both!

np Bhorok's answer is more elegant and easier, but I thought you might need a long winded explanation and there's often more than one way to swing a cat I guess. Hope it helped.

- #8

- 740

- 13

(x^2)/((x^2+1)^2)is(1/2)(arctan(x)-(x/x^2+1))

In class, we've seen the steps to solve this integral, but I don't understand certain steps..

Can someone explain me how to solve this integral, step by step?

since you have the answer, take its derivative & work backwards. that's how to figure it out. just don't show anyone your rough work :tongue2:

- #9

uart

Science Advisor

- 2,776

- 9

Sure, no problem, here are the steps:

Integral((x^2)/((x^2+1)^2)dx)

= (1/2)*Integral(x d(-1/(x^2+1))

= (-1/2)*(x/(x^2+1)) -(1/2)*Integral(1/(x^2+1)dx)

= -(1/2)*(x/(x^2+1))+(arctan(x))/2

Last step is the answer

(The answer I said earlier was wrong, this is the correct answer:(1/2)*(x/(x^2+1))-(arctan(x))/2)

Thank you

No the original answer was correct, you dropped a minus sign in the first line of this derivation.

Share: