integrate log base 10 x / x dx
i don't understand this
To integrate the function log base 10 (x) / x, you can use integration by parts.
Let's rewrite the function as log(x) / (x * log(10)):
∫(log(x) / (x * log(10))) dx
Let's assign:
u = log(x) (u-substitution)
dv = dx / (x * log(10)) (choose dv such that its integral is easier to calculate)
du = (1/x) dx (differentiate u)
v = (1 / log(10)) ln(x) (integrate dv)
Now we can apply the integration by parts formula:
∫(u * dv) = u * v - ∫(v * du)
Applying this formula:
∫(log(x) / (x * log(10))) dx
= (1 / log(10)) * log(x) * (1 / log(10)) ln(x) - ∫[(1 / log(10)) ln(x) * (1/x) dx]
Simplifying the expression:
= (1 / log(10))^2 * log(x) * ln(x) - ∫(1 / (x * log(10))) dx
Now, we need to solve the integral ∫(1 / (x * log(10))) dx. This is a simple integration.
∫(1 / (x * log(10))) dx
= (1 / log(10)) ∫(1 / x) dx
= (1 / log(10)) ln|x| + C (where C is the constant of integration)
Putting it all together, the final result is:
∫(log(x) / (x * log(10))) dx
= (1 / log(10))^2 * log(x) * ln(x) - (1 / log(10)) ln|x| + C
That's the step-by-step process of integrating the given function.
To integrate the function log base 10 of x divided by x with respect to x, there are different methods you can use. Let's explore one approach called integration by parts.
Integration by parts is a technique based on the product rule of differentiation. It states that for two functions u(x) and v(x), the integral of their product can be expressed as the product of two functions minus the integral of the product of their derivatives:
∫ u(x) v'(x) dx = u(x) v(x) - ∫ u'(x) v(x) dx
Applying this technique to the given integral, we can choose u(x) = log base 10 of x and v'(x) = 1/x. To find u'(x) and v(x), we differentiate u(x) and integrate v'(x):
u'(x) = 1/(x ln(10)) (using the logarithmic differentiation rule)
v(x) = ln(x) (the antiderivative of 1/x)
Using the formula for integration by parts, we have:
∫ (log base 10 of x / x) dx = u(x) v(x) - ∫ u'(x) v(x) dx
= log base 10 of x * ln(x) - ∫ (1/(x ln(10))) * ln(x) dx
Now, we need to evaluate the second integral on the right side of the equation. This integral can be simplified using a substitution.
Letting u = ln(x), we have du = (1/x) dx. Substituting these values into our integral:
∫ (1/(x ln(10))) * ln(x) dx = ∫ du
= u + C
= ln(x) + C
Finally, substituting this result back into our original equation:
∫ (log base 10 of x / x) dx = log base 10 of x * ln(x) - ln(x) + C
= ln(x) * (log base 10 of x - 1) + C
So the definite integral of log base 10 of x divided by x with respect to x is ln(x) * (log base 10 of x - 1) plus a constant of integration, C.
log(10)x = log(10)e*log(e)x
= 0.4342945 ln x
So, integrate 0.4342945 ln x /x
It will help to know that the integral of (ln x)/x is (ln x)^2/ 2.