What is the integral of x-1+4sin(2x) ?

The integral of x-1+4sin(2x) is (x^2)/2-x-2cos(2x)+C

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integral of x-1+4sin(2x)

\int x - 1 + 4 sin (2 x) d x

\frac{x ^{2}}{2} - x - 2 cos (2 x) + C

Solution steps

\int x - 1 + 4 sin (2 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x d x - \int 1 d x + \int 4 sin (2 x) d x

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

\int 4 sin (2 x) d x = - 2 cos (2 x)

= \frac{x ^{2}}{2} - x - 2 cos (2 x)

Add a constant to the solution

= \frac{x ^{2}}{2} - x - 2 cos (2 x) + C

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What is the integral of x-1+4sin(2x) ?
The integral of x-1+4sin(2x) is (x^2)/2-x-2cos(2x)+C