# Thread: Substitution Rule Integral Questions

1. ## 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.

2. 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

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.

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: $\frac 12 \ln \bigg| \ln |x^2 + 1| \bigg| + C$

3. 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.

4. 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.

5. 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.