Integration substitution.

[rowdy3]

Use substitution to find each indefinite integral.
∫ (square root 2 + lnx) / x ; dx
I did
u=2+ln(x)
then differentiate both sides to get
du=0+dx/x
∫ (square root 2 + lnx) / x ; dx
∫ (square root u) du
∫ u^0.5 du
u^(3/2)/(3/2)+c
=(2+ln(x))^(3/2) +c
Is the answer right? Thanks.

 

[LCKurtz]

Use substitution to find each indefinite integral.
∫ (square root 2 + lnx) / x ; dx
I did
u=2+ln(x)
then differentiate both sides to get
du=0+dx/x
∫ (square root 2 + lnx) / x ; dx
∫ (square root u) du
∫ u^0.5 du
u^(3/2)/(3/2)+c
=(2+ln(x))^(3/2) +c
Is the answer right? Thanks.

It was until you left out something in the last line.

 

[rowdy3]

(2/3)*[2+lnx]^(3/2)+C. Is that right?

 

[Dick]

(2/3)*[2+lnx]^(3/2)+C. Is that right?

Yes, it is.