What is the integral of 2x+8sin(x) ?

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

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integral of 2x+8sin(x)

\int 2 x + 8 sin (x) d x

x^{2} - 8 cos (x) + C

Solution steps

\int 2 x + 8 sin (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 2 x d x + \int 8 sin (x) d x

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

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

= x^{2} - 8 cos (x)

Add a constant to the solution

= x^{2} - 8 cos (x) + C

\int (sec^{2} (x) - 2) d x

\int_{0}^{2} \frac{2}{2 - x} d x

\int_{0}^{1} e^{- 8 x} d x

x \to \infty lim \frac{( e ^{\frac{15}{x}} - 1 )}{\frac{15}{x}}

(s + \frac{3}{2})^{2}

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