# Thread: Can't seem to get this right (integrals)

1. ## Can't seem to get this right (integrals)

I'm using a new method learned in class and cannot apply it properly. I made the table of integration and then simplified but still got it wrong. Where did I go wrong?

2. I can't understand this method.The solution is this:

3. the method is called tabular integration and it requires that you
take antiderivatives in the $e^{3x}$ column ...

$e^{3x}$

$\frac{1}{3}e^{3x}$

$\frac{1}{9}e^{3x}$

$\frac{1}{27}e^{3x}$

4. I couldn't get the other problem so I'm trying a similar one. I am applying the tabular table of integration method correctly but did not get the correct result. Here is my second attempt.

5. your writing leaves much to be desired ...

$7x^2 - 4x$ ............... $e^{5x}$

$14x - 4$ .................. $\frac{1}{5}e^{5x}$

$14$ ........................ $\frac{1}{25}e^{5x}$

$0$ ......................... $\frac{1}{125}e^{5x}$

$(7x^2-4x) \cdot \frac{1}{5}e^{5x} - (14x-4) \cdot \frac{1}{25}e^{5x} + 14 \cdot \frac{1}{125}e^{5x} + C$

$e^{5x}\left(\frac{7x^2-4x}{5} - \frac{14x-4}{25} + \frac{14}{125}\right) + C$

your result is the same as mine ... why do you say it's wrong? (you did forget the constant of integration)

6. Is it okay to do what I did with the denominator and turn the 1/5, 1/25 terms into 1/125, multiplied the numerators by the correct amount and then factored out 1/125?

7. Originally Posted by solidstatemath
Is it okay to do what I did with the denominator and turn the 1/5, 1/25 terms into 1/125, multiplied the numerators by the correct amount and then factored out 1/125?
yes, it's ok.

8. I'm going to have to see my teacher about this. I input my answer and it said it was wrong, then input yours and it said it was wrong only to say the correct answer was the one I gave in the first place.

Thanks for your help, time to break some computers!