Question

Can someone explain to me how the indefinite integral of squareroot of x over x is 2 squareroot of x?

Answer 1

This is my function \[\int\limits_{}^{}\frac{ 1+\sqrt{x} }{ x }dx\]

Answer 2

I get how 1/x is lnx but I'm not sure how wolfram alpha is getting, \[2\]

Answer 3

\[2\sqrt{x}\]

Answer 4

What did you type in wolfram alpha?

Answer 5

integrate (1+x^(1/2))/(x)

Answer 6

you're integrating sqrt(x)/x=1/sqrt(x) for that second part

Answer 7

1/sqrt(x)=x^(-1/2)
divide by -1/2+1 and take x to the power of -1/2+1 according to power rule and you get
2x^(1/2)=2sqrt(x)

Answer 8

@JerJason ?

Answer 9

sorry i had to leave for a moment.

Answer 10

How is sqrt(x)/x=1/sqrt(x)?