Why Does My Integral Substitution Result in a Logarithm?

[Stevo6754]

Homework Statement

\int2x^3/2x^2+1

Homework Equations

None

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 haven't 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?

 

[MisterMan]

EDIT: At first sight it looked like the one was separate from the 2x^2, sorry I realized 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?

 

[Stevo6754]

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 I am a little baffled.

 

[jegues]

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.

 

[HallsofIvy]

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

 

[Leptos]

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.

 

[rs1n]

Homework Statement

\int2x^3/2x^2+1

Homework Equations

None

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 haven't 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?

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?

 

[Char. Limit]

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?

 

[HallsofIvy]

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

 

[Leptos]

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?

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

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

Wolfram Alpha.

 

[Dickfore]

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

<br /> 2 x^{3} \div (2x^{2} + 1) = x - \frac{x}{2x^{2} + 1}<br />

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

 

[rs1n]

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.

Im stuck, problem is this looks like its going to be a log, but we haven't 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?

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