What is the integral of (cos(x)+1/9 x) ?

The integral of (cos(x)+1/9 x) is sin(x)+(x^2)/(18)+C

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integral of (cos(x)+1/9 x)

\int (cos (x) + \frac{1}{9} x) d x

sin (x) + \frac{x ^{2}}{18} + C

Solution steps

\int cos (x) + \frac{1}{9} 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 cos (x) d x + \int \frac{1}{9} x d x

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

\int \frac{1}{9} x d x = \frac{x ^{2}}{18}

= sin (x) + \frac{x ^{2}}{18}

Add a constant to the solution

= sin (x) + \frac{x ^{2}}{18} + C

x \to 1 lim (\frac{2 - 7 x}{x + 6})

5 lo g_{10} (x)

y^{^{''}} - 7 y^{^{'}} = 49 t, y (0) = 7, y^{^{'}} (0) = 0

x \to 0 lim (\frac{1}{x ^{2}})

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e^{- 4 x}, y (0) = - 2, y^{^{'}} (0) = 1

What is the integral of (cos(x)+1/9 x) ?
The integral of (cos(x)+1/9 x) is sin(x)+(x^2)/(18)+C