What is the integral of (sin(x)+1)' ?

The integral of (sin(x)+1)' is x(sin(x)+1)'-xcos(x)+sin(x)+C

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integral of (sin(x)+1)'

\int (sin (x) + 1)^{'} d x

x (sin (x) + 1)' - x cos (x) + sin (x) + C

Solution steps

\int (sin (x) + 1)^{'} d x

Apply Integration By Parts

= x (sin (x) + 1)^{^{'}} - \int - x sin (x) d x

\int - x sin (x) d x = x cos (x) - sin (x)

= x (sin (x) + 1)^{^{'}} - (x cos (x) - sin (x))

Simplify

= x (sin (x) + 1)^{^{'}} - x cos (x) + sin (x)

Add a constant to the solution

= x (sin (x) + 1)^{^{'}} - x cos (x) + sin (x) + C

What is the integral of (sin(x)+1)' ?
The integral of (sin(x)+1)' is x(sin(x)+1)'-xcos(x)+sin(x)+C