# Substitution Rule Integral Questions

• Jul 15th 2008, 11:40 PM
SportfreundeKeaneKent
Substitution Rule Integral Questions
I have a few questions I'm unsure about, the first one is a indefinite integral:

1.) Find: antiderivative sqrt(1+lnx)*lnx*1/x dx

I got 2u^(5/2)/5 - 2u^(3/2)/3 for this one where u=1+lnx

The next two are definite

2.) antiderivative 2x/sqrt(4x^2 + 3) dx where the interval is between 0 and 2

For this one, I got 1/4*antiderivative 1/sqrtu where u=4x^2 + 3 and the interval is between 3 and 19. The problem is that when I sub in the x's back in for u and do the math (the interval changes back to 0/2, then I get a completely different answer for both.

3.) antiderivative x/(x^2+1)ln(x^2+1)

For this one, I get an antiderivative of 1/2ln|u| between 0.69 and 1.61 when u = ln(x^2+1). Doing the math, I get 0.42 as the answer.

But when I sub the x's back in, the antiderivative changes to 1/2ln|x^2+1| between 1 and 2 and I get 0.46 for my answer. So either I'm making a mistake or the two aren't equal.
• Jul 16th 2008, 12:20 AM
Jhevon
Quote:

Originally Posted by SportfreundeKeaneKent
I have a few questions I'm unsure about, the first one is a indefinite integral:

1.) Find: antiderivative sqrt(1+lnx)*lnx*1/x dx

I got 2u^(5/2)/5 - 2u^(3/2)/3 for this one where u=1+lnx

correct. but write your answer in terms of x and don't forget the +C

Quote:

The next two are definite

2.) antiderivative 2x/sqrt(4x^2 + 3) dx where the interval is between 0 and 2

For this one, I got 1/4*antiderivative 1/sqrtu where u=4x^2 + 3 and the interval is between 3 and 19. The problem is that when I sub in the x's back in for u and do the math (the interval changes back to 0/2, then I get a completely different answer for both.
sorry, don't see your problem here, i got the same answer both ways. if you are going to back-substitute the x expression, there is no point in changing the limit in the first place.

Quote:

3.) antiderivative x/(x^2+1)ln(x^2+1)

For this one, I get an antiderivative of 1/2ln|u| between 0.69 and 1.61 when u = ln(x^2+1). Doing the math, I get 0.42 as the answer.

But when I sub the x's back in, the antiderivative changes to 1/2ln|x^2+1| between 1 and 2 and I get 0.46 for my answer. So either I'm making a mistake or the two aren't equal.
you lost me. where'd 0.69 and 1.61 come from. how did the limits become 1 and 2?

also, if i read your problem correctly (you should use parentheses), i think you mean $\int \frac x{(x^2 + 1)\ln (x^2 + 1)}~dx$ as opposed to $\int \frac x{x^2 + 1} \cdot \ln (x^2 + 1)~dx$

then, the integral is: $\frac 12 \ln \bigg| \ln |x^2 + 1| \bigg| + C$
• Jul 16th 2008, 12:24 AM
mr fantastic
Quote:

Originally Posted by SportfreundeKeaneKent
I have a few questions I'm unsure about, the first one is a indefinite integral:

1.) Find: antiderivative sqrt(1+lnx)*lnx*1/x dx

I got 2u^(5/2)/5 - 2u^(3/2)/3 for this one where u=1+lnx

The next two are definite

2.) antiderivative 2x/sqrt(4x^2 + 3) dx where the interval is between 0 and 2

For this one, I got 1/4*antiderivative 1/sqrtu where u=4x^2 + 3 and the interval is between 3 and 19.

Mr F says: Now calculate this definite integral.

The problem is that when I sub in the x's back in for u and do the math (the interval changes back to 0/2, then I get a completely different answer for both.

Mr F says: Why would you substitute back?? This is totally unnecessary. It wastes time and increases the probabilty of making an error and getting the wrong answer.

3.) antiderivative x/(x^2+1)ln(x^2+1)

For this one, I get an antiderivative of 1/2ln|u| between 0.69 and 1.61 when u = ln(x^2+1). Doing the math, I get 0.42 as the answer.

But when I sub the x's back in, the antiderivative changes to 1/2ln|x^2+1| between 1 and 2 and I get 0.46 for my answer. So either I'm making a mistake or the two aren't equal.

For questions 2.) and 3.) it will be easier to help you if you show all your working.
• Jul 16th 2008, 12:01 PM
SportfreundeKeaneKent
For the third one, I'm getting the same antiderivative as you. When I compute it between 1 and 2, I get an answer of 0.46.

For the second one, I got an answer of -1.31.
• Jul 16th 2008, 03:24 PM
mr fantastic
Quote:

Originally Posted by SportfreundeKeaneKent
For the third one, I'm getting the same antiderivative as you. When I compute it between 1 and 2, I get an answer of 0.46.

For the second one, I got an answer of -1.31.

The answer for the second one is positive not negative. And it's possible to give an exact answer: $\frac{\sqrt{19} - \sqrt{3}}{2}$.

The answer for the third one, correct to two decimal places, is 0.42. Again, it's possible to give an exact answer: $\frac{1}{2} \, \ln \frac{\ln 5}{\ln 2}$. You're probably making a data entry error when attempting to get an approximate answer.