What is the integral of 8cos(x)+3x-8 ?

The integral of 8cos(x)+3x-8 is 8sin(x)+(3x^2)/2-8x+C

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integral of 8cos(x)+3x-8

\int 8 cos (x) + 3 x - 8 d x

8 sin (x) + \frac{3 x ^{2}}{2} - 8 x + C

Solution steps

\int 8 cos (x) + 3 x - 8 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 8 cos (x) d x + \int 3 x d x - \int 8 d x

\int 8 cos (x) d x = 8 sin (x)

\int 3 x d x = \frac{3 x ^{2}}{2}

\int 8 d x = 8 x

= 8 sin (x) + \frac{3 x ^{2}}{2} - 8 x

Add a constant to the solution

= 8 sin (x) + \frac{3 x ^{2}}{2} - 8 x + C

y = ln (\frac{1}{( x - 2 ) ^{3}})

- 5 e^{4 t}

\frac{\partial}{\partial x} (x sin (2 x^{2} y))

\int \frac{1}{t 4 t ^{2} - 1} d t

y^{^{''}} - y = x^{2} e^{x}

What is the integral of 8cos(x)+3x-8 ?
The integral of 8cos(x)+3x-8 is 8sin(x)+(3x^2)/2-8x+C