Help with log integration/differntiation

[myanmar]

I'm having trouble on these two problems. Can anyone give me a step by step explanation on how to do them (or one of them)?

7. Find f'(x), if
f(x) = $$\left|\frac {x^2\left((3x + 2)^{\frac {1}{3}}\right)}{(2x - 3)^3}\right|$$

13. Perform the integration:
(a) $$\int(x-1)e^{-x^2+2x}dx$$
(b) $$\int\frac{1}{x}e^{-2\log_3(x)}dx$$

 

[HallsofIvy]

I'm having trouble on these two problems. Can anyone give me a step by step explanation on how to do them (or one of them)?

7. Find f'(x), if
f(x) = $$\left|\frac {x^2\left((3x + 2)^{\frac {1}{3}}\right)}{(2x - 3)^3}\right|$$

This has nothing to do with a logarithm! Do it as two separate problems:
$$f(x)= x^2(3x+2)^{1/3}(2x-3)^{-3}$$ if $$x\ge 0$$
$$f(x)= -x^2(3x+2)^{1/3}(2x-3)^{-3}$$ if x< 0
Use the chain rule and product rule

13. Perform the integration:
(a) $$\int(x-1)e^{-x^2+2x}dx$$

The derivative of -x²+ 2x is -2x+2= -2(x-1). Use the substitution u= x²-2x.

(b) $$\int\frac{1}{x}e^{-2\log_3(x)}dx$$

Finally a problem involving a lograrithm!

-2 log₃(x)= log₃(x³)= -2ln(x)/ln(3) so
$$e^{-2 log_3(x)}= 3^ln(x)$$. Let u= ln(x).

 

[Dick]

This has nothing to do with a logarithm! Do it as two separate problems:
$$f(x)= x^2(3x+2)^{1/3}(2x-3)^{-3}$$ if $$x\ge 0$$
$$f(x)= -x^2(3x+2)^{1/3}(2x-3)^{-3}$$ if x< 0

? If g(x)=x^2*(3x+2)^(1/3)*(2x-3)^(-3), (so f(x)=|g(x)|), you need to find the regions where g(x) is positive or negative. Not where x is positive or negative. Find the x values where g(x) is 0 or undefined. Those are the endpoints of the intervals where g(x) has uniform sign.

 

[myanmar]

Thanks for the help so far,

trying 13 i get
a. u=-x^2 + 2x
du = -2(x-1)dx
int((x-1)/(-2(x-1)) e^u) su
int( -1/2 e^u du) = -1/2 e^u u' + c = (x-1)(e^(-x^2+2x)) + c

b.
so, I have int(1/x e^(-2lnx/ln3)) dx
u=lnx
du = 1/x dx
int (e^(-2u/ln3)) du = -2/ln3 e^(-2u/ln3) + c = -2/ln3 e^(-2lnx/ln3) + c = -2/ln3 e^(-2 log_3 x) + c

is this correct?
i'm not quite sure where i'd put the absolute value signs, but i know i need them.

 

[rock.freak667]

Thanks for the help so far,

trying 13 i get
a. u=-x^2 + 2x
du = -2(x-1)dx
int((x-1)/(-2(x-1)) e^u) su
int( -1/2 e^u du) = -1/2 e^u u' + c = (x-1)(e^(-x^2+2x)) + c /QUOTE]

The last line is wrong. Remember that $\int e^x dx=e^x +c$

 

[myanmar]

Based on Halls of Ivy / Dick 's help, I get this for 1
$$\frac{-(x)(12x^2-26x-27)}{(2x-3)^{-4}(3x+2)^{-2/3}}$$ if $${x} > \frac{3}{2}$$or $${x} \leq \frac{-2}{3}$$
$$\frac{-(x)(12x^2-26x-27)}{(2x-3)^{-4}(3x+2)^{-2/3}}$$ if $$\frac{3}{2} < {x} < \frac{-2}{3}$$

is this right?

 

[Dick]

Based on Halls of Ivy / Dick 's help, I get this for 1
$$\frac{-(x)(12x^2-26x-27)}{(2x-3)^{-4}(3x+2)^{-2/3}}$$ if $${x} > \frac{3}{2}$$or $${x} \leq \frac{-2}{3}$$
$$\frac{-(x)(12x^2-26x-27)}{(2x-3)^{-4}(3x+2)^{-2/3}}$$ if $$\frac{3}{2} < {x} < \frac{-2}{3}$$

is this right?

No... There should be a sign difference between the two expressions on different intervals. And that derivative doesn't look right. The powers of the factors in the denominator should be positive. And I don't get the same quadratic you have in the numerator.

 

[myanmar]

Oh, sorry.
I meant

$$\frac{-(x)(12x^2-26x-27)}{(2x-3)^{4}(3x+2)^{2/3}}$$ if $${x} > \frac{3}{2}$$or $${x} \leq \frac{-2}{3}$$
$$\frac{(x)(12x^2-26x-27)}{(2x-3)^{4}(3x+2)^{2/3}}$$ if $$\frac{3}{2} < {x} < \frac{-2}{3}$$

 

[Dick]

That's better. But I still don't get the same coefficients you do on the quadratic part.