First make a substitution and then use integration by parts to evaluate. The integral of (x)(ln(x+3))dx What do you substitute first?
let y = x+3
then dy =dx
(y-3) ln y dy
= y ln y dy - 3 ln y dy
= y ln y dy - 3/y
let u = ln y then dv = y dy
so v = (1/2)y^2 and du = dy/y
so
(1/2)y^2 ln y -(1/2)ydy -3/y
(1/2)y^2 ln y - (1/4)y^2 - 3/y
but remember y = x + 3 :)
screw dis
To evaluate the integral of (x)(ln(x+3))dx, we can make a substitution and then use integration by parts.
We first need to choose an appropriate substitution to simplify the integral. Let's substitute u = ln(x+3).
To find du, we differentiate both sides of the equation u = ln(x+3) with respect to x:
du/dx = 1/(x+3) * (d/dx)(x+3) = 1/(x+3)
Now, we can rewrite the integral in terms of u and du:
∫ x ln(x+3) dx = ∫ u du
Next, we can integrate the simplified integral ∫ u du using the power rule of integration:
∫ u du = (1/2)u^2 + C
Finally, we need to substitute back the original variable x:
= (1/2)(ln(x+3))^2 + C
So, the integral of (x)(ln(x+3))dx is (1/2)(ln(x+3))^2 + C.