Question

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

Answer 1

`I=2/9*x^(3/2)[3ln(2x)-2]+C`

Explanation:

`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`