How do you integrate #int sqrtx ln 2x dx # using integration by parts?

Question

How do you integrate #int sqrtx   ln 2x  dx  # using integration by parts?

Evelyn Agard · Accepted Answer

#I=2/9*x^(3/2)[3ln(2x)-2]+C# #color(red)(I=intu*vdx=u*intvdx-int(u^'*intvdx)dx)# #I=intx^(1/2)*ln(2x)dx,# #Take,u=ln(2x)and v=x^(1/2)# #rArru^'=1/(2x)*2=1/xand intvdx=(x^(3/2))/(3/2)=2/3*x^(3/2)+c# #I=ln(2x)*2/3*x^(3/2)-int1/x*2/3*x^(3/2)dx# #I=2/3*x^(3/2)ln(2x)-2/3intx^(1/2)dx# #I=2/3*x^(3/2)ln(2x)-2/3*(x^(3/2))/(3/2)+C# #I=2/3*x^(3/2)ln(2x)-4/9*x^(3/2)+C# #I=2/9*x^(3/2)[3ln(2x)-2]+C#

Aurora Alvarado · Answer

undefined To integrate ∫sqrt(x) ln(2x) dx using integration by parts, let u = ln(2x) and dv = sqrt(x) dx. Then differentiate u to find du and integrate dv to find v. Apply the integration by parts formula:

∫u dv = uv - ∫v du

Calculate du and v:

du = (1/x) dx
v = (2/3)x^(3/2)

Now apply the integration by parts formula:

∫sqrt(x) ln(2x) dx = (2/3)x^(3/2) ln(2x) - ∫(2/3)x^(3/2) * (1/x) dx

Simplify the integral:

∫sqrt(x) ln(2x) dx = (2/3)x^(3/2) ln(2x) - (2/3)∫x^(1/2) dx

Integrate the simplified integral:

∫x^(1/2) dx = (2/3)x^(3/2)

Therefore, the final result is:

∫sqrt(x) ln(2x) dx = (2/3)x^(3/2) ln(2x) - (4/9)x^(3/2) + C