- #1

Juan Pablo

- 40

- 0

I just learned how to integrate through substitution and I was challenged by my teacher with an apparently easy problem but I'm really struggling with it.

He said he will give me an F if I don't solve it for tomorrow, I guess this is what I get by being the one who always understand in class

Exposed below

[tex]\int (x^2+1)^2 dx[/tex]

[tex]x^2+1=u[/tex]

[tex]\frac{du}{2x}=dx[/tex]

[tex]x=\sqrt{u-1}[/tex]

[tex]\int u^2(\frac{du}{2x}) = [/tex]

[tex]\int u^2(\frac{du}{2\sqrt{u-1}}) = [/tex]

[tex]\frac{1}{2}\int\frac{du}{u^{-2}\sqrt{u-1}} = [/tex]

Now I finally integrate

[tex]

\frac{1}{2} ln(u^{-2}\sqrt{u-1})[/tex]

[tex]\frac{1}{2} ln(\frac{\sqrt{u-1}}{u^2})[/tex]

[tex]\frac{1}{2} ln(\frac{x}{(x^2+1)^2})[/tex]But if I integrate it solving the binomial first I get a totally different answer and If I derive my result I don't get [tex](x^2+1)^2[/tex]

Anybody?

He said he will give me an F if I don't solve it for tomorrow, I guess this is what I get by being the one who always understand in class

Exposed below

## Homework Statement

[tex]\int (x^2+1)^2 dx[/tex]

## Homework Equations

[tex]x^2+1=u[/tex]

[tex]\frac{du}{2x}=dx[/tex]

[tex]x=\sqrt{u-1}[/tex]

## The Attempt at a Solution

[tex]\int u^2(\frac{du}{2x}) = [/tex]

[tex]\int u^2(\frac{du}{2\sqrt{u-1}}) = [/tex]

[tex]\frac{1}{2}\int\frac{du}{u^{-2}\sqrt{u-1}} = [/tex]

Now I finally integrate

[tex]

\frac{1}{2} ln(u^{-2}\sqrt{u-1})[/tex]

[tex]\frac{1}{2} ln(\frac{\sqrt{u-1}}{u^2})[/tex]

[tex]\frac{1}{2} ln(\frac{x}{(x^2+1)^2})[/tex]But if I integrate it solving the binomial first I get a totally different answer and If I derive my result I don't get [tex](x^2+1)^2[/tex]

Anybody?

Last edited: