# Thread: Major Integration Help

1. ## Major Integration Help

A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!

2. Originally Posted by dillonmhudson
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!
1. Are you sure the numerator isn't $\displaystyle x^2 + 3x + 2$???

3. Originally Posted by dillonmhudson
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!
1. Long divide. Then the integrand becomes $\displaystyle x - 4 - \frac{2}{x + 1}$. Much easier.

4. You thinking about factoring? Yeah I'm positive, I double checked that.

5. Originally Posted by dillonmhudson
You thinking about factoring? Yeah I'm positive, I double checked that.
Worth a shot.

You'll have to long divide in that case.

6. Originally Posted by dillonmhudson
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!
3. Remember that $\displaystyle \tan{5x} = \frac{\sin{5x}}{\cos{5x}}$.

Make the substitution $\displaystyle u = \cos{5x}$.

7. Originally Posted by dillonmhudson
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!
For 5. You'll have to long divide as well.

$\displaystyle \frac{x^2 - 1}{x + 1} = x - 1 - \frac{1}{x + 1}$.

8. Originally Posted by dillonmhudson
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!
$\displaystyle 1.{\text{ }}\int {\frac{{{x^2} - 3x + 2}}{{x + 1}}} dx = \int {\frac{{\left( {x + 1} \right)x - \left( {4x - 2} \right)}}{{x + 1}}} dx = \int {\left[ {x - \frac{{4x - 2}}{{x + 1}}} \right]} dx =$

$\displaystyle = \frac{{{x^2}}}{2} - 2\int {\frac{{2x - 1}}{{x + 1}}} dx = \frac{{{x^2}}}{2} - 2\int {\frac{{2x + 2 - 3}}{{x + 1}}} dx = \frac{{{x^2}}} {2} - 2\int {\left[ {2 - \frac{3}{{x + 1}}} \right]} dx =$

$\displaystyle = \frac{{{x^2}}}{2} - 4\int {dx} + 6\int {\frac{{dx}}{{x + 1}}} = \frac{{{x^2}}}{2} - 4x + 6\ln \left( {x + 1} \right) + C.$

9. Originally Posted by DeMath

$\displaystyle = \frac{{{x^2}}}{2} - 4\int {dx} + 6\int {\frac{{dx}}{{x + 1}}} = \frac{{{x^2}}}{2} - 4x + 6\ln \left( {x + 1} \right) + C.$
Not quite on the $\displaystyle \ln(x+1),$ something is missing...

10. This $\displaystyle \ln \left| {x + 1} \right|$?

Thanks!

11. Originally Posted by Prove It
1. Long divide. Then the integrand becomes $\displaystyle x - 4 - \frac{2}{x + 1}$. Much easier.
I think the integrand should be $\displaystyle x - 4 + \frac{6}{x + 1}$ after the long division.