How Do You Solve an Integral Using Substitution?

rowdy3
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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.
 
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rowdy3 said:
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.
 
(2/3)*[2+lnx]^(3/2)+C. Is that right?
 
Last edited:
rowdy3 said:
(2/3)*[2+lnx]^(3/2)+C. Is that right?

Yes, it is.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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