USE INTEGRATION BY PARTS TO FIND EACH INTEGRAL
ƪsquare rootx lnx dx
Please identify the School Subject clearly. Is this Calculus? Then, it should say so.
If you tried to cut and paste, it doesn't work here. Please type it all out as this is not at all clear.
Sra
let u = lnx
du/dx = (1/x)
du = (1/x) dx
let dv = x^(1/2) dx
v = (2/3)x^(3/2)
[int]√x lnx dx = .....
how about you giving it a try, let me know what you got
wolframalpha dot com
Type:
integrate sqrt(x)lnx dx
and click option =
then:
Indefinite integral: Show steps
To find the integral of √xlnx dx using integration by parts, we need to assign the different parts of the integrand to be the "u" and "dv" terms.
Let's choose u = ln(x) and dv = √x dx.
Now we need to find du and v.
To find du, we differentiate u with respect to x.
du/dx = 1/x.
Therefore, du = (1/x) dx.
To find v, we integrate dv with respect to x.
Integrating √x dx gives us v = (2/3)x^(3/2).
Now, we can apply the integration by parts formula:
∫u dv = uv - ∫v du
Using the formula, we will have:
∫√xlnx dx = ∫u dv
∫√xlnx dx = u*v - ∫v du
∫√xlnx dx = ln(x) * (2/3)x^(3/2) - ∫(2/3)x^(3/2) * (1/x) dx
Simplifying the above expression, we have:
∫√xlnx dx = (2/3)x^(3/2)ln(x) - (2/3) ∫x^(1/2) dx.
The integral of x^(1/2) can be easily evaluated using the power rule of integration:
∫x^(1/2) dx = (2/3) x^(3/2).
Finally, substituting back into the original expression, we get:
∫√xlnx dx = (2/3)x^(3/2)ln(x) - (2/3) (2/3) x^(3/2) + C
Simplifying further, we obtain our final result:
∫√xlnx dx = (2/3)x^(3/2)ln(x) - (4/9) x^(3/2) + C
where C is the constant of integration.