What is the integral of (cos(x)+1/6 x) ?

The integral of (cos(x)+1/6 x) is sin(x)+(x^2)/(12)+C

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integral of (cos(x)+1/6 x)

\int (cos (x) + \frac{1}{6} x) d x

sin (x) + \frac{x ^{2}}{12} + C

Solution steps

\int cos (x) + \frac{1}{6} 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}{6} x d x

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

\int \frac{1}{6} x d x = \frac{x ^{2}}{12}

= sin (x) + \frac{x ^{2}}{12}

Add a constant to the solution

= sin (x) + \frac{x ^{2}}{12} + C

\int - \frac{x ^{3}}{1 - x ^{2}} d x

\frac{d}{d x} (e^{a x^{5}})

x \to 0 lim (2 x (1 - ln (x)))

\frac{d}{d x} (arcsin (x^{9}))

d er i v a t i v e f (x) = 6 x^{2} - 4 x

What is the integral of (cos(x)+1/6 x) ?
The integral of (cos(x)+1/6 x) is sin(x)+(x^2)/(12)+C