# Homework Help: Integral substitution trouble

1. Jul 25, 2010

### Stevo6754

1. The problem statement, all variables and given/known data

$$\int2x^3/2x^2+1$$

2. Relevant equations
None

3. The attempt at a solution
I used substitution

u = 2x^2+1
du/dt = 4x
dt = du/4x

$$\int(2x^3/u)du/4x$$

cancel out the x to get
$$\int(2x^2/u)du/4$$

solve for 2x^2
u = 2x^2 + 1
2x^2 = u - 1

$$\int((u-1)/u)du/4$$

$$\int(1/4)(1-1/u)du$$

Im stuck, problem is this looks like its going to be a log, but we havent done logs in class so the answer he is looking for shouldn't be a log, am I going about this the wrong way or is his question wrong?

2. Jul 25, 2010

### MisterMan

Re: Integrate

EDIT: At first sight it looked like the one was separate from the 2x^2, sorry I realised my mistake when I scrolled down.

It looks like a simple log integration to me. What other integration concepts have you studied in class recently?

Last edited: Jul 25, 2010
3. Jul 25, 2010

### Stevo6754

Re: Integrate

substitution, definite integrals, fundamental theorem of calculus.. Pretty much the basics of integration, we just started areas between curves and volumes. He has never done log functions for any integration problems so Im a little baffled.

4. Jul 25, 2010

### jegues

Re: Integrate

Just simplify the integral.

$$\int\left( \frac{2x^{3}}{2x^{2}} +1 \right)dx = \int (x+1)dx = \int xdx + \int dx$$

Should be childs play now.

5. Jul 26, 2010

### HallsofIvy

Re: Integrate

Is the problem
$$\int\left(\frac{2x^3}{2x^2}+1\right)dx$$

which is what jeques assumed and what you wrote or is it $\int 2x^3/(2x^2+ 1) dx$?

If it is
$$\int\frac{2x^3}{2x^2+ 1}dx$$
then divide first:
$$\frac{2x^3}{2x^2+ 1}= x- \frac{x}{2x^2+1}$$

$$\int\frac{2x^3}{2x^2+ 1}dx= \int x dx- \int\frac{x}{2x^2+ 1}dx$$

6. Jul 26, 2010

### Leptos

Re: Integrate

First you must factor out the constant, and then you must use long division. Next you just separate the integral into pieces and substitute. Remember to reverse-substitute so your final computation is in terms of the original variable.

Note: I didn't see any problems that required long division until the beginning of Calculus II(but for many textbooks it's covered in Calculus I). Regardless, you should know long division of polynomials from Pre-Calculus.

7. Jul 27, 2010

### rs1n

Re: Integrate

This approach is correct -- just remember to resubstitute u=2x^2+1. Except you should have du/dx = 4x and not du/dt = 4x. However, I do not see how you can avoid logarithms. Did the question strictly forbid the use of logs?

8. Jul 27, 2010

### Char. Limit

Re: Integrate

This might be a bad place, but when I type the equation into W-A (NOT the integral, the integrand), the integral section adds 1/4 to the final answer, even though there's already a +C. Checking "show steps" doesn't reveal a reason... can anyone explain?

9. Jul 27, 2010

### HallsofIvy

Re: Integrate

I have no idea what "W-A" is.

10. Jul 27, 2010

### Leptos

Re: Integrate

It's for restricted values for the family of functions.
If you expand (2x2 + 1)/4 you will get x2/2 + 1/4

Wolfram Alpha.

11. Jul 27, 2010

### Dickfore

Re: Integrate

Hint. The polynomial in the numerator is of higher degree (3rd) than the polynomial in the denominator (2nd). So, do the long division first:

$$2 x^{3} \div (2x^{2} + 1) = x - \frac{x}{2x^{2} + 1}$$

Now, you have two integrals. The first one is elementary and the second one can be integrated with the substitution you had suggested.

12. Jul 27, 2010

### rs1n

Re: Integrate

I think we're all overlooking the actual question. The original post already had what could be considered the beginning of a correct solution (assuming the most natural interpretation of the typos). If I am interpreting the question correctly, the original poster was looking either for 1) a solution that does NOT involve a logarithm as part of the antiderivative or 2) an explanation why such a solution would not be possible.

Everyone else seems to be giving alternate or equivalent solutions that seem to sidestep the real issue in the original post.