MHB Help with an integral (substitution method)

[fernlund]

Hello! I need help to integrate the square root of:

p x² + x⁴

by the method of substitution, p is just a constant. I've been trying for a long time but I can't get it right. I know the answer and it still doesn't help. Thanks! (sorry, I don't know the Latex stuff)

 

[MarkFL]

I would begin by factoring under the radical in the integrand:

$$\int\sqrt{x^2\left(x^2+p\right)}\,dx$$

What do you suppose the next step should be?

 

[fernlund]

I would begin by factoring under the radical in the integrand:

$$\int\sqrt{x^2\left(x^2+p\right)}\,dx$$

What do you suppose the next step should be?

Thank you so much. Actually, that's about as far as I've gotten. Then I was thinking I should subtitute

u = $$p + x^2 $$

and maybe break out the $$\sqrt{x^2} $$ but that would give me $$\left|x\right|$$ right? Or rather, this is wrong, right?

$$\int x \sqrt{p + x^2} $$

I think I'm missing something obvious.

 

[MarkFL]

Yes, we would have:

$$\int|x|\sqrt{x^2+p}\,dx$$

I would first treat $x$ as if is is positive, and concern myself with the sign of $x$ at the end. So next we write:

$$\int x\sqrt{x^2+p}\,dx$$

So, using the substitution you suggest:

$$u=x^2+p$$

What differential do we need?

 

[fernlund]

Yes, we would have:

$$\int|x|\sqrt{x^2+p}\,dx$$

I would first treat $x$ as if is is positive, and concern myself with the sign of $x$ at the end. So next we write:

$$\int x\sqrt{x^2+p}\,dx$$

So, using the substitution you suggest:

$$u=x^2+p$$

What differential do we need?

Well my next step would be

$$\d{u}{x} = 2x$$ and

x = $$\ \d{u}{x} \frac{1}{2}$$

which would lead to

$$\ \frac{1}{2} \int \sqrt{u} du \implies $$

$$\ \frac{1}{3} {u}^{\frac{3}{2}} + C $$

but that's not correct, so I'm still missing something.

 

[MarkFL]

Yes, I would write:

$$u=x^2+p\,\therefore\,du=2x\,dx$$

and if we write the integral as:

$$\frac{1}{2}\int\sqrt{x^2+p}\cdot2x\,dx$$

then applying the substitution, we get:

$$\frac{1}{2}\int u^{\frac{1}{2}}\,du$$

Now, applying the power rule for integration, we have:

$$\frac{1}{2}\int u^{\frac{1}{2}}\,du=\frac{1}{2}\left(\frac{u^{\frac{3}{2}}}{\dfrac{3}{2}}\right)+C=\frac{1}{3}u^{\frac{3}{2}}+C$$

So, next we want to back-substitute for $u$, and then multiply by the sign of $x$, which can be written as $$\frac{|x|}{x}$$...what do you get?

 

[fernlund]

Yes, I would write:

$$u=x^2+p\,\therefore\,du=2x\,dx$$

and if we write the integral as:

$$\frac{1}{2}\int\sqrt{x^2+p}\cdot2x\,dx$$

then applying the substitution, we get:

$$\frac{1}{2}\int u^{\frac{1}{2}}\,du$$

Now, applying the power rule for integration, we have:

$$\frac{1}{2}\int u^{\frac{1}{2}}\,du=\frac{1}{2}\left(\frac{u^{\frac{3}{2}}}{\dfrac{3}{2}}\right)+C=\frac{1}{3}u^{\frac{3}{2}}+C$$

So, next we want to back-substitute for $u$, and then multiply by the sign of $x$, which can be written as $$\frac{|x|}{x}$$...what do you get?

Oh yeah, that's correct! Thank you so much, it was the sign of x that did it. But then I have another question.

Actually, what I want to do is calculate the original, definite integral from x = -1 to = 2 which is a bit problematic since the antiderivate which we just calculated isn't continuous at x = 0.

I could calculate the integral by looking at two intervals, from -1 to 0 and then 0 to 2, and use the limits at x = 0, but shouldn't I be able to do it directly with the u-substitution and changing the interval from [-1,2] to [28,31] (u = x^2 +p)?

If I do that I don't get the right answer.

 

[MarkFL]

I would integrate in terms of $x$, and treat it as two integrals as you suggest, two improper integrals...

 

[fernlund]

I would integrate in terms of $x$, and treat it as two integrals as you suggest, two improper integrals...

OK, thank you so much! It really means a lot to me that you took your time :) Is it even possible to calculate it as a definite integral in the u-form?

 

[MarkFL]

OK, thank you so much! It really means a lot to me that you took your time :) Is it even possible to calculate it as a definite integral in the u-form?

The $u$-form does not consider the sign of $x$...that's why I would use the $x$-form. :D