Solving for indefinite integral using u-substitution follows:
`int f(g(x))*g'(x) dx = int f(u) du` where we let` u = g(x)` .
In this case, it is stated that to let u be the denominator of integral which means let:
`u = 1+sqrt(2x).`
This can be rearrange into `sqrt(2x) = u -1`
Finding the derivative of u : `du = 1/sqrt(2x) dx`
Substituting `sqrt(2x)= u-1` into `du = 1/sqrt(2x)dx` becomes:
`du = 1/(u-1)dx`
Rearranged into `(u-1)...
Solving for indefinite integral using u-substitution follows:
`int f(g(x))*g'(x) dx = int f(u) du` where we let` u = g(x)` .
In this case, it is stated that to let u be the denominator of integral which means let:
`u = 1+sqrt(2x).`
This can be rearrange into `sqrt(2x) = u -1`
Finding the derivative of u : `du = 1/sqrt(2x) dx`
Substituting `sqrt(2x)= u-1` into `du = 1/sqrt(2x)dx` becomes:
`du = 1/(u-1)dx`
Rearranged into `(u-1) du =dx`
Applying u-substitution using` u = 1+sqrt(2x) ` and `(u-1)du = du` :
`int 1/(1+sqrt(2x)) dx = int (u-1)/u *du`
Express into two separate fractions:
`int (u-1)/u *du = int ( u/u -1/u)du`
` = int (1 - 1/u)du`
Applying `int (f(x) -g(x))dx = int f(x) dx - int g(x) dx` :
`int (1 - 1/u)du = int 1 du - int 1/udu`
`= u - ln|u| +C`
Substitute `u = 1+sqrt(2x) ` to the `u - ln|u| +C` :
`u - ln|u| +C =1+sqrt(2x) -ln|1+sqrt(2x) |+C`
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