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{9}{x ^{6}} d x

\int 3 x^{2} (x^{3} + 5)^{4} d x

\int \frac{sin ( x )}{1 + 3 cos ( x )} d x

\int \frac{12 - x}{x} d 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