# Math Help - Definite Integrals

1. ## Definite Integrals

Hello!
I have tried to solve these two integrals, but I get the wrong answer.

The first one is 4/x^2 - 4 (upper limit is 1, lower limit 0).
The correct answer is -ln2, but I get 2arctan(x/2) + c.

The next one is ln^2 x/2x (upper limit e^2, lower limit 1).
I am not sure what the correct answer is, but either
a) 4/3 b) 1/4 c) 3/5 d) 7/6 e) 4/9

I would appreciate your help,
best wishes
Amine

2. Please reply showing your steps, so we can try to help you find any errors. Thank you!

3. Originally Posted by Aminekhadir
Hello!
I have tried to solve these two integrals, but I get the wrong answer.

The first one is 4/x^2 - 4 (upper limit is 1, lower limit 0).
The correct answer is -ln2, but I get 2arctan(x/2) + c.

The next one is ln^2 x/2x (upper limit e^2, lower limit 1).
I am not sure what the correct answer is, but either
a) 4/3 b) 1/4 c) 3/5 d) 7/6 e) 4/9

I would appreciate your help,
best wishes
Amine
It will be much easier to adress your troubles and misconceptions (and you have several of each) if you post all your work. Also, it would help if your equations were less ambiguous. I assume by 4/x^2 - 4 you mean 4/(x^2 - 4) and by ln^2 x/2x you mean (ln^2 x)/(2x) ....

4. Originally Posted by Aminekhadir
Hello!
I have tried to solve these two integrals, but I get the wrong answer.

The first one is 4/x^2 - 4 (upper limit is 1, lower limit 0).
The correct answer is -ln2, but I get 2arctan(x/2) + c.
I presume you mean 4/(x^2-4), not (4/x^2)- 4. "arctan" is the integral of dx/(x^2+1), not dx/(x^2-4). To integrate that, write 1/(x^2-4) as 1/((x-2)(x+2)) and use "partial fractions". That is, find A and B such that 4/(x^2- 4)= A/(x-2)+ B(x+2) and integrate those separately.

The next one is ln^2 x/2x (upper limit e^2, lower limit 1).
I am not sure what the correct answer is, but either
a) 4/3 b) 1/4 c) 3/5 d) 7/6 e) 4/9
Use the simple substitution u= ln(x).

I would appreciate your help,
best wishes
Amine

5. for the second integral it is 4/3 . integral ( 1 -> e^2) { ln^2x / 2x } = 1/6(ln^3 x) from 1 - > e^2 => 1/6 ( 8 - 0 ) = 8/6 = 4/3

6. for the first use partial fraction , note you are getting arc tan and thats wrong becoz the derivative of arctan is in the demoniator + , not -

7. Originally Posted by Aminekhadir
Hello!
I have tried to solve these two integrals, but I get the wrong answer.

The first one is 4/x^2 - 4 (upper limit is 1, lower limit 0).
The correct answer is -ln2, but I get 2arctan(x/2) + c.

$\frac{4}{x^2-4}=\frac{1}{x-2}-\frac{1}{x+2}$ and thus this a a logarithmic integral: $\ln\left|\frac{x-2}{x+2}\right|_0^1=\ln\frac{1}{3}-\ln 1=-\ln 3$ , so imo both answers are incorrect.

The next one is ln^2 x/2x (upper limit e^2, lower limit 1).
I am not sure what the correct answer is, but either
a) 4/3 b) 1/4 c) 3/5 d) 7/6 e) 4/9

$\int\limits_1^{e^2}\frac{\ln^2x}{2x}=\frac{1}{6}\l eft[\ln^3x\right]_1^{e^2}=\frac{1}{6}(8-0)=\frac{4}{3}$

Tonio

I would appreciate your help,
best wishes
Amine
.