Expert: Paul Klarreich - 2/20/2007

Question
I had to do 100 problems for a test on integration and could not figure out 10 of them..... I already got the test back and received a 90... hence I got them wrong... My teacher does not provide the answers and i'd really like to know how to do them...
All of these are FIND THE INDEFINITE INTEGRAL

1.) t^3 divided by t^4 + 1 dt

2.) x^6 * lnx dx

3.) t divided by t^2 - 1

4.) (2t+1)(t^2 + t - 1)^8

5.) (t+e^4t)

6.) ln(x^3)

7.) 1 divided by 1+e^x

8.) x divided by the square root of 1 + x

THESE LAST TWO ARE SOLVE THE GIVEN INITIAL VALUE PROBLEM

9.) dy/dt = t - te^t , y(0)=-1

10.) dy/dt = 1 divided by t^2 + 3t + 2, y(0)=0

Answer
Questioner:   Richard Hearon
Category:  Calculus
 
Subject:  Calculus: Integration
Question:  I had to do 100 problems for a test on integration and could not figure out 10 of them..... I already got the test back and received a 90... hence I got them wrong... My teacher does not provide the answers

>> WELL, HE SHOULD! HE'S GETTING PAID FOR THAT, AND NOBODY SAID THAT TEACHING DIDN'T INVOLVE SOME WORK.

and i'd really like to know how to do them...
All of these are FIND THE INDEFINITE INTEGRAL

1.) t^3 divided by t^4 + 1 dt

2.) x^6 * lnx dx

3.) t divided by t^2 - 1

4.) (2t+1)(t^2 + t - 1)^8

5.) (t+e^4t)

6.) ln(x^3)

7.) 1 divided by 1+e^x

8.) x divided by the square root of 1 + x

THESE LAST TWO ARE SOLVE THE GIVEN INITIAL VALUE PROBLEM

9.) dy/dt = t - te^t , y(0)=-1

10.) dy/dt = 1 divided by t^2 + 3t + 2, y(0)=0
....................................................
Hi, Richard,

This is a lot of problems, so I'm not going to work them out for you.  I'll indicate the method you should choose, and leave the rest to you.

1.) t^3 divided by t^4 + 1 dt

Do a substitution:  u = t^4 + 1.  ln(t^4 + 1) will come up in the answer.

2.) x^6 * lnx dx

Do integration by parts (IBP),  u = ln x,  dv = x^6 dx. (the ln x will differentiate to 1/x, and the rest should be routine.)

3.) t divided by t^2 - 1

Let  u = t^2 - 1, you should get a ln(t^2 - 1) in the answer.

4.) (2t+1)(t^2 + t - 1)^8

Let  u = t^2 + t - 1,  du = (2t + 1)dt

5.) (t+e^4t)

Integrate each term separately.  For  e^4t, let u = 4t

6.) ln(x^3) = 3 ln x, then use IBP.

7.) 1 divided by 1+e^x

Let  u = 1+e^x, du = e^x dx = (u - 1)dx.

8.) x divided by the square root of 1 + x

Let u = sqrt(1 + x), a rationalizing substitution.  Then  x = u^2 - 1 and you should be able to do the rest.

THESE LAST TWO ARE SOLVE THE GIVEN INITIAL VALUE PROBLEM

9.) dy/dt = t - te^t , y(0)=-1

Integrate each term.  Do  t e^t as IBP, u = t, du = e^t dt

10.) dy/dt = 1 divided by t^2 + 3t + 2, y(0)=0

Write  t^2 + 3t + 2 = (t + 2)(t + 1), then do Partial Fractions. (I hate them, too, but...)

If you get stuck on one or two, let me know.