find the indefinite integral of
(1/x)*(e^(-2log[ 3,x])) dx
where 3 is base of the log
i cannot even figure out where to start :( please help. thank you
log_3(x) = lnx/ln3
e^(-2log_3(x))
= e^(-2lnx/ln3)
= (e^(lnx))^(-2/ln3)
= x^(-2/ln3)
so you have
∫(1/x) x^(-2/ln3) dx
= ∫x^(-2/ln3-1) dx
= 1/(-2/ln3) x^(-2/ln3)
or
-ln3/2 x^(-2/ln3)
To find the indefinite integral of the given expression, we can make use of a few differentiation and integration rules. Let's break down the problem step by step:
1. Start by simplifying the expression inside the parentheses: e^(-2log[3,x]).
- Since the natural logarithm (ln) is the inverse of the exponential function, we can rewrite the expression as e^(-log[3,x]^2).
- Apply the logarithmic identity: a^(-b) = 1/a^b. Therefore, e^(-log[3,x]^2) can be rewritten as 1/(e^(log[3,x]^2)).
2. Convert the expression with a logarithmic base of 3 to a natural logarithmic base (ln):
- Use the change of base formula: log[b,x] = log[a,x] / log[a,b]. Applying this to the given expression, we have log[3,x] = ln(x) / ln(3).
- Now, substituting this value back into our expression, we have 1/(e^((ln(x)/ln(3))^2)).
3. Simplify the expression further:
- Since the square of any natural logarithm is equal to the natural logarithm squared, the expression can be simplified as 1/(e^(ln(x)^2/ln(3)^2)).
- The expression can be rewritten as 1/(e^(ln(x^2)/ln(3)^2)).
4. Apply the property of logarithms:
- Recall that e^(ln(a)) = a. Using this property, we can rewrite the expression as 1/(x^2/ln(3)^2).
- Simplifying further, we have ln(3)^2/x^2.
5. Finally, integrate the expression:
- The indefinite integral of ln(3)^2/x^2 with respect to x is ln(3)^2 * (1/x) + C, where C is the constant of integration.
So, the indefinite integral of (1/x) * (e^(-2log[3,x])) is ln(3)^2 * (1/x) + C.
To find the indefinite integral of the given function, we can simplify the expression first.
Using the properties of logarithms, we can rewrite the term within the exponential function:
e^(-2log[ 3,x]) = e^(log[3,x^(-2)]) = x^(-2).
So, the integral becomes:
∫((1/x) * (e^(-2log[ 3,x]))) dx = ∫((1/x) * x^(-2)) dx.
Next, we can simplify the expression (1/x) * x^(-2) by combining the fractions:
(1/x) * x^(-2) = x^(-1) * x^(-2) = x^(-1-2) = x^(-3).
Now we can find the indefinite integral of x^(-3):
∫ x^(-3) dx
To integrate x^(-3), we can add 1 to the exponent and multiply by the reciprocal of the new exponent:
∫ x^(-3) dx = (x^(-2)) / -2 + C
where C is the constant of integration.
Therefore, the indefinite integral of the given function is:
∫((1/x) * (e^(-2log[ 3,x]))) dx = (x^(-2)) / -2 + C.