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

1 Answer
maganbhai P.
Mar 7, 2018

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

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