- #1
rowdy3
- 33
- 0
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.
∫ (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.