What is the integral of (1-1/x)cos(x-ln(x)) ?

The integral of (1-1/x)cos(x-ln(x)) is sin(x-ln(x))+C

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integral of (1-1/x)cos(x-ln(x))

\int (1 - \frac{1}{x}) cos (x - ln (x)) d x

sin (x - ln (x)) + C

Solution steps

\int (1 - \frac{1}{x}) cos (x - ln (x)) d x

Apply u-substitution

: \int cos (u) d u

= \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= sin (u)

Substitute back

u = x - ln (x)

= sin (x - ln (x))

Add a constant to the solution

= sin (x - ln (x)) + C

\int \frac{1}{- x ^{2} - x + 2} d x

\int ln^{2} (y) d y

\int x^{4} x d x

\int x arcsin (4 x) d x

\int \frac{x ^{4}}{1 + x ^{5}} d x

What is the integral of (1-1/x)cos(x-ln(x)) ?
The integral of (1-1/x)cos(x-ln(x)) is sin(x-ln(x))+C